title stringlengths 1 100 | titleSlug stringlengths 3 77 | Java int64 0 1 | Python3 int64 1 1 | content stringlengths 28 44.4k | voteCount int64 0 3.67k | question_content stringlengths 65 5k | question_hints stringclasses 970
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🚀Beats 99.17% || DFS Recursive & Iterative || Euler Path Intuition || Commented Code🚀 | reconstruct-itinerary | 1 | 1 | # Problem Description\nThe task is **reconstructing** an itinerary using a list of airline tickets.\nEach ticket is represented as a pair of departure and arrival airports, denoted as `tickets[i] = [fromi, toi]`.\n**The goal** is to create an **ordered** itinerary, starting with the airport `JFK` (the departure point),... | 45 | You are given a list of airline `tickets` where `tickets[i] = [fromi, toi]` represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.
All of the tickets belong to a man who departs from `"JFK "`, thus, the itinerary must begin with `"JFK "`. If there are multiple... | null |
✔96.90% Beats C++ ✨📈|| Hard--->Easy 😇|| Easy to Understand👁 || #Beginner😉😎 | reconstruct-itinerary | 1 | 1 | # Intuition\n- The problem requires finding a valid itinerary for a given list of flight tickets. \n- The itinerary must start from "JFK" (John F. Kennedy International Airport) and visit all airports exactly once. \n- This problem can be solved using depth-first search (DFS) on a directed graph, treating each airport ... | 16 | You are given a list of airline `tickets` where `tickets[i] = [fromi, toi]` represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.
All of the tickets belong to a man who departs from `"JFK "`, thus, the itinerary must begin with `"JFK "`. If there are multiple... | null |
Python 80% solution with descriptive code and comment about decision | reconstruct-itinerary | 0 | 1 | # Code\n```\nclass Solution:\n def findItinerary(self, tickets: List[List[str]]) -> List[str]:\n num_tickets = len(tickets)\n\n conns = defaultdict(list)\n for i, (src, dest) in enumerate(tickets):\n conns[src].append((dest, i))\n\n # Sorted the tickets so that the first fully-... | 1 | You are given a list of airline `tickets` where `tickets[i] = [fromi, toi]` represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.
All of the tickets belong to a man who departs from `"JFK "`, thus, the itinerary must begin with `"JFK "`. If there are multiple... | null |
PYTHON | EASILY EXPLAINED SOLUTION | reconstruct-itinerary | 0 | 1 | \n\n# Approach\n\nThis code defines a `findItinerary` function that performs the following steps:\n\n1. Create a dictionary `graph` to represent the graph of tickets, where each airport is a key, and the values are lists of destination airports sorted in reverse order to ensure that smaller lexical order airports are v... | 1 | You are given a list of airline `tickets` where `tickets[i] = [fromi, toi]` represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.
All of the tickets belong to a man who departs from `"JFK "`, thus, the itinerary must begin with `"JFK "`. If there are multiple... | null |
Simply simple Python Solution - Using stack for dfs - with comments | reconstruct-itinerary | 0 | 1 | \tclass Solution:\n def findItinerary(self, tickets: List[List[str]]) -> List[str]:\n graph = {}\n # Create a graph for each airport and keep list of airport reachable from it\n for src, dst in tickets:\n if src in graph:\n graph[src].append(dst)\n else:\n ... | 190 | You are given a list of airline `tickets` where `tickets[i] = [fromi, toi]` represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.
All of the tickets belong to a man who departs from `"JFK "`, thus, the itinerary must begin with `"JFK "`. If there are multiple... | null |
332: Solution with step by step explanation | reconstruct-itinerary | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nWe first create a heap for each source vertex, containing the destinations in lexicographical order. Then, during the DFS traversal, we pop the smallest destination from the heap instead of sorting the destinations each time... | 6 | You are given a list of airline `tickets` where `tickets[i] = [fromi, toi]` represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.
All of the tickets belong to a man who departs from `"JFK "`, thus, the itinerary must begin with `"JFK "`. If there are multiple... | null |
[Python] Simple Solution | reconstruct-itinerary | 0 | 1 | ```\nclass Solution:\n def findItinerary(self, tickets: List[List[str]]) -> List[str]:\n ## RC ##\n ## APPROACH : GRAPH DFS ##\n ## EDGE CASE : [["JFK","KUL"],["JFK","NRT"],["NRT","JFK"]]\n \n\t\t## TIME COMPLEXITY : O(N) ##\n\t\t## SPACE COMPLEXITY : O(N) ##\n \n def dfs(ci... | 13 | You are given a list of airline `tickets` where `tickets[i] = [fromi, toi]` represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.
All of the tickets belong to a man who departs from `"JFK "`, thus, the itinerary must begin with `"JFK "`. If there are multiple... | null |
Python3 DFS Solution | reconstruct-itinerary | 0 | 1 | ```\nclass Solution:\n def findItinerary(self, tickets: List[List[str]]) -> List[str]:\n graph = {}\n \n tickets.sort(key=lambda x: x[1])\n\n for u, v in tickets:\n if u in graph:\n graph[u].append(v)\n else:\n graph[u] = [v]\n \n... | 14 | You are given a list of airline `tickets` where `tickets[i] = [fromi, toi]` represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.
All of the tickets belong to a man who departs from `"JFK "`, thus, the itinerary must begin with `"JFK "`. If there are multiple... | null |
Python Easy Solution | increasing-triplet-subsequence | 0 | 1 | \n\n# Code\n```\n\nclass Solution:\n def increasingTriplet(self, nums: List[int]) -> bool:\n first = second = float(\'inf\') \n for n in nums: \n if n <= first: \n first = n\n elif n <= second:\n second = n\n else:\n return T... | 63 | Given an integer array `nums`, return `true` _if there exists a triple of indices_ `(i, j, k)` _such that_ `i < j < k` _and_ `nums[i] < nums[j] < nums[k]`. If no such indices exists, return `false`.
**Example 1:**
**Input:** nums = \[1,2,3,4,5\]
**Output:** true
**Explanation:** Any triplet where i < j < k is valid.
... | null |
python 3 || 6 lines, one-pass, w/explanation || T/M: 98%/50% | increasing-triplet-subsequence | 0 | 1 | The plan here is iterate through nums, and place each element in the least position (first thought third) for which it qualifies, should it qualify. If we find a third, we succeed (True), otherwise we fail (False).\n\n```\nclass Solution:\n def increasingTriplet(self, nums: list[int]) -> bool:\n \n fir... | 49 | Given an integer array `nums`, return `true` _if there exists a triple of indices_ `(i, j, k)` _such that_ `i < j < k` _and_ `nums[i] < nums[j] < nums[k]`. If no such indices exists, return `false`.
**Example 1:**
**Input:** nums = \[1,2,3,4,5\]
**Output:** true
**Explanation:** Any triplet where i < j < k is valid.
... | null |
Python3 math based, non-optimized O(n) space + time solution | increasing-triplet-subsequence | 0 | 1 | # Intuition\nI really had no idea how to solve this in O(n) time, so I messed around with it for a couple days without success before taking a step back and attempting to find a mathematical solution on pen and paper first.\n\nWe start with the following condition:\n\n$$x_i < x_j < x_k$$\n\nFor array of length $$N$$ th... | 3 | Given an integer array `nums`, return `true` _if there exists a triple of indices_ `(i, j, k)` _such that_ `i < j < k` _and_ `nums[i] < nums[j] < nums[k]`. If no such indices exists, return `false`.
**Example 1:**
**Input:** nums = \[1,2,3,4,5\]
**Output:** true
**Explanation:** Any triplet where i < j < k is valid.
... | null |
99.76% faster in Python||cpp || java -TC=O(n), Sc(1) || Easy to Understand | increasing-triplet-subsequence | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code utilizes a greedy approach to find an increasing triplet subsequence by keeping track of the two smallest elements f and s. If it encounters a larger element, it confirms the existence of an increasing triplet subsequence. Otherw... | 16 | Given an integer array `nums`, return `true` _if there exists a triple of indices_ `(i, j, k)` _such that_ `i < j < k` _and_ `nums[i] < nums[j] < nums[k]`. If no such indices exists, return `false`.
**Example 1:**
**Input:** nums = \[1,2,3,4,5\]
**Output:** true
**Explanation:** Any triplet where i < j < k is valid.
... | null |
[Python] 2 solutions: Right So Far, One pass - O(1) Space - Clean & Concise | increasing-triplet-subsequence | 0 | 1 | **\u2714\uFE0F Solution 1: Build Max Right So Far and Max Left So Far**\n- Let `maxRight[i]` denote the maximum number between `nums[i+1], nums[i+2],...nums[n-1]`.\n```python\nclass Solution:\n def increasingTriplet(self, nums: List[int]) -> bool:\n n = len(nums)\n maxRight = [0] * n # maxRight[i] is ... | 138 | Given an integer array `nums`, return `true` _if there exists a triple of indices_ `(i, j, k)` _such that_ `i < j < k` _and_ `nums[i] < nums[j] < nums[k]`. If no such indices exists, return `false`.
**Example 1:**
**Input:** nums = \[1,2,3,4,5\]
**Output:** true
**Explanation:** Any triplet where i < j < k is valid.
... | null |
334: Solution with step by step explanation | increasing-triplet-subsequence | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Define a function increasingTriplet that takes a list of integers nums as input and returns a boolean value indicating if there exists a triplet of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k].\n\... | 16 | Given an integer array `nums`, return `true` _if there exists a triple of indices_ `(i, j, k)` _such that_ `i < j < k` _and_ `nums[i] < nums[j] < nums[k]`. If no such indices exists, return `false`.
**Example 1:**
**Input:** nums = \[1,2,3,4,5\]
**Output:** true
**Explanation:** Any triplet where i < j < k is valid.
... | null |
Increasing Triplets Python Solution | increasing-triplet-subsequence | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFind two numbers, where the first is smaller than the second. If we can find a third number that\'s bigger than both, we\'ve found a triplet.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nInitialize our `smallest... | 3 | Given an integer array `nums`, return `true` _if there exists a triple of indices_ `(i, j, k)` _such that_ `i < j < k` _and_ `nums[i] < nums[j] < nums[k]`. If no such indices exists, return `false`.
**Example 1:**
**Input:** nums = \[1,2,3,4,5\]
**Output:** true
**Explanation:** Any triplet where i < j < k is valid.
... | null |
Very Easy || 100% || Fully Explained || Java, C++, Python, JavaScript, Python3 | increasing-triplet-subsequence | 1 | 1 | # **PROBLEM STATEMENT:**\nGiven an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.\n# **Example:**\n# Input: nums = [2,1,5,0,4,6]\n# Output: true\n# Explanation: The triplet (3, 4, 5) is valid bec... | 21 | Given an integer array `nums`, return `true` _if there exists a triple of indices_ `(i, j, k)` _such that_ `i < j < k` _and_ `nums[i] < nums[j] < nums[k]`. If no such indices exists, return `false`.
**Example 1:**
**Input:** nums = \[1,2,3,4,5\]
**Output:** true
**Explanation:** Any triplet where i < j < k is valid.
... | null |
Python easy Sol. O(n)time complexity | increasing-triplet-subsequence | 0 | 1 | ```\nclass Solution:\n def increasingTriplet(self, nums: List[int]) -> bool:\n first=second=float(\'inf\')\n for i in nums:\n if i<=first:\n first=i\n elif i<=second:\n second=i\n else:\n return True\n return False | 16 | Given an integer array `nums`, return `true` _if there exists a triple of indices_ `(i, j, k)` _such that_ `i < j < k` _and_ `nums[i] < nums[j] < nums[k]`. If no such indices exists, return `false`.
**Example 1:**
**Input:** nums = \[1,2,3,4,5\]
**Output:** true
**Explanation:** Any triplet where i < j < k is valid.
... | null |
Simple Python Solution 100% Accepted | TC O(n) SC O(1) | increasing-triplet-subsequence | 0 | 1 | \n# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(1)$$\n\n# Code\n```\nclass Solution:\n def increasingTriplet(self, arr: List[int]) -> bool:\n i = j = float(\'inf\')\n for num in arr:\n if num <= i:\n i = num\n elif num <= j:\n ... | 3 | Given an integer array `nums`, return `true` _if there exists a triple of indices_ `(i, j, k)` _such that_ `i < j < k` _and_ `nums[i] < nums[j] < nums[k]`. If no such indices exists, return `false`.
**Example 1:**
**Input:** nums = \[1,2,3,4,5\]
**Output:** true
**Explanation:** Any triplet where i < j < k is valid.
... | null |
335: Solution with step by step explanation | self-crossing | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe algorithm checks three possible cases for each line of the path, starting from the 3rd line. If any of the cases is true, the function returns True, meaning the path crosses itself. If none of the cases is true after loo... | 7 | You are given an array of integers `distance`.
You start at the point `(0, 0)` on an **X-Y plane,** and you move `distance[0]` meters to the north, then `distance[1]` meters to the west, `distance[2]` meters to the south, `distance[3]` meters to the east, and so on. In other words, after each move, your direction chan... | null |
Python solution: 3 ways to self cross | self-crossing | 0 | 1 | # Intuition\r\n<!-- Describe your first thoughts on how to solve this problem. -->\r\nSimulating the process would take up too much memory and time. We need a geometrical solution\r\n# Approach\r\n<!-- Describe your approach to solving the problem. -->\r\nLogically break down the possibilities of self cross: it could c... | 1 | You are given an array of integers `distance`.
You start at the point `(0, 0)` on an **X-Y plane,** and you move `distance[0]` meters to the north, then `distance[1]` meters to the west, `distance[2]` meters to the south, `distance[3]` meters to the east, and so on. In other words, after each move, your direction chan... | null |
Python 95% | No math, Just space for time, make it an easy problem | self-crossing | 0 | 1 | # Intuition\r\nSince the problem scale is not very large, why not exchange space for time?\r\n\r\n# Approach\r\nkeep track of every single points we pass throught\r\n\r\n# Complexity\r\n- Time complexity:\r\n$$O(nm)$$\r\n\r\n- Space complexity:\r\n$$O(nm)$$\r\n\r\n# Code\r\n```\r\nfrom operator import sub\r\nMOVE = [(0... | 1 | You are given an array of integers `distance`.
You start at the point `(0, 0)` on an **X-Y plane,** and you move `distance[0]` meters to the north, then `distance[1]` meters to the west, `distance[2]` meters to the south, `distance[3]` meters to the east, and so on. In other words, after each move, your direction chan... | null |
A State Machine Approach | self-crossing | 0 | 1 | # Intuition\nA none-intersecting path is either:\n\n- an out-growing spiral (spiral-out), or\n- an in-growing spiral (spiral-in), or\n- an out-growing spiral connected with an in-growing one.\n# Approach\nWe use a state-machine approach where the states are the phase of current spiral we are in:\n\n- GROW: we are spira... | 0 | You are given an array of integers `distance`.
You start at the point `(0, 0)` on an **X-Y plane,** and you move `distance[0]` meters to the north, then `distance[1]` meters to the west, `distance[2]` meters to the south, `distance[3]` meters to the east, and so on. In other words, after each move, your direction chan... | null |
A State Machine Approach | self-crossing | 0 | 1 | # Intuition\n\nA none-intersecting path is either:\n1. an out-growing spiral (spiral-out), or\n2. an in-growing spiral (spiral-in), or\n3. an out-growing spiral connected with an in-growing one.\n\n# Approach\n\nWe use a state-machine approach where the states are the phase of current spiral we are in:\n\n- **GROW**: w... | 0 | You are given an array of integers `distance`.
You start at the point `(0, 0)` on an **X-Y plane,** and you move `distance[0]` meters to the north, then `distance[1]` meters to the west, `distance[2]` meters to the south, `distance[3]` meters to the east, and so on. In other words, after each move, your direction chan... | null |
Expressive solution 🙂 | self-crossing | 0 | 1 | Very bad drawing but how i approached this question\n\n\n# Code\n```\nclass Solution:\n def isSelfCrossing(self, distance: List[int]) -> bool:\n """\n 1st way to cross NWSE , East meets nort... | 0 | You are given an array of integers `distance`.
You start at the point `(0, 0)` on an **X-Y plane,** and you move `distance[0]` meters to the north, then `distance[1]` meters to the west, `distance[2]` meters to the south, `distance[3]` meters to the east, and so on. In other words, after each move, your direction chan... | null |
Bad | self-crossing | 0 | 1 | # Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(n)$$\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n$$O(1)$$\n\n# Code\n```\nclass Solution:\n def isSelfCrossing(self, distance: List[int]) -> bool:\n if len(distance) <= 3:\n ... | 0 | You are given an array of integers `distance`.
You start at the point `(0, 0)` on an **X-Y plane,** and you move `distance[0]` meters to the north, then `distance[1]` meters to the west, `distance[2]` meters to the south, `distance[3]` meters to the east, and so on. In other words, after each move, your direction chan... | null |
Solution | self-crossing | 1 | 1 | ```C++ []\nstatic const auto _ = []{ return ios_base::sync_with_stdio(false), 0; }();\n\nclass Solution {\npublic:\n bool isSelfCrossing(vector<int>& distance)\n {\n int r[7]{};\n int i = 0;\n for (; i < distance.size(); ++i)\n {\n __builtin_memmove(&r[1], &r[0], sizeof(int)... | 0 | You are given an array of integers `distance`.
You start at the point `(0, 0)` on an **X-Y plane,** and you move `distance[0]` meters to the north, then `distance[1]` meters to the west, `distance[2]` meters to the south, `distance[3]` meters to the east, and so on. In other words, after each move, your direction chan... | null |
Spatial Reasoning | Explained & Commented | self-crossing | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWhen we consider what a crossing is, it is the moment when we have a planar cross on both the horizontal and vertical axis of their number lines within the last 6 total steps. For this, you need to record 5 current steps and the one you a... | 0 | You are given an array of integers `distance`.
You start at the point `(0, 0)` on an **X-Y plane,** and you move `distance[0]` meters to the north, then `distance[1]` meters to the west, `distance[2]` meters to the south, `distance[3]` meters to the east, and so on. In other words, after each move, your direction chan... | null |
Best O(n) solution | self-crossing | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe solution to this problem involves checking if the path of the person will cross itself. One way to do this is to check if the path intersects any previous lines that it has traced.\n# Approach\n<!-- Describe your approach to solving t... | 0 | You are given an array of integers `distance`.
You start at the point `(0, 0)` on an **X-Y plane,** and you move `distance[0]` meters to the north, then `distance[1]` meters to the west, `distance[2]` meters to the south, `distance[3]` meters to the east, and so on. In other words, after each move, your direction chan... | null |
Python 93.5% || very concise and understandable solution || 10 lines of code | self-crossing | 0 | 1 | ```\n\n\n# Written by : Dhruv_Vavliya\n\n# just think about 4,5,6 --length arrays \ndef self_crossing(x):\n for i in range(3,len(x)):\n\n if x[i-3]>=x[i-1] and x[i]>=x[i-2]:\n return True\n \n if i>=4:\n if x[i-3] == x[i-1] and x[i-2]<=(x[i-4]+x[i]):\n return... | 3 | You are given an array of integers `distance`.
You start at the point `(0, 0)` on an **X-Y plane,** and you move `distance[0]` meters to the north, then `distance[1]` meters to the west, `distance[2]` meters to the south, `distance[3]` meters to the east, and so on. In other words, after each move, your direction chan... | null |
Python - Top 80% Speed - O(n) time, O(1) space - Dimensionless Coordinates | self-crossing | 0 | 1 | **Python - Top 80% Speed - O(n) time, O(1) space - Dimensionless Coordinates**\n\nThe code below uses Dimensionless Coordinates (p,q) to detect collisions with walls. The solution is fairly efficient/valid, but the conversions between Coordinate Systems can be tedious.\n\n```\ndef point_intersection(qa0,qb0,pc0,walls):... | 1 | You are given an array of integers `distance`.
You start at the point `(0, 0)` on an **X-Y plane,** and you move `distance[0]` meters to the north, then `distance[1]` meters to the west, `distance[2]` meters to the south, `distance[3]` meters to the east, and so on. In other words, after each move, your direction chan... | null |
Intuitive Python3 | HashMap | 95% Time & Space | O(N*W^2) | palindrome-pairs | 0 | 1 | Please upvote if it helps!\n```\nclass Solution:\n def palindromePairs(self, words: List[str]) -> List[List[int]]:\n backward, res = {}, []\n for i, word in enumerate(words):\n backward[word[::-1]] = i\n\n for i, word in enumerate(words):\n \n if word in backward... | 63 | You are given a **0-indexed** array of **unique** strings `words`.
A **palindrome pair** is a pair of integers `(i, j)` such that:
* `0 <= i, j < words.length`,
* `i != j`, and
* `words[i] + words[j]` (the concatenation of the two strings) is a palindrome.
Return _an array of all the **palindrome pairs** of_ `... | null |
EASY PYTHON SOLUTION || USING MAPPING | palindrome-pairs | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | You are given a **0-indexed** array of **unique** strings `words`.
A **palindrome pair** is a pair of integers `(i, j)` such that:
* `0 <= i, j < words.length`,
* `i != j`, and
* `words[i] + words[j]` (the concatenation of the two strings) is a palindrome.
Return _an array of all the **palindrome pairs** of_ `... | null |
Best and easy to understand code | palindrome-pairs | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | You are given a **0-indexed** array of **unique** strings `words`.
A **palindrome pair** is a pair of integers `(i, j)` such that:
* `0 <= i, j < words.length`,
* `i != j`, and
* `words[i] + words[j]` (the concatenation of the two strings) is a palindrome.
Return _an array of all the **palindrome pairs** of_ `... | null |
Python with Comments ✅ | palindrome-pairs | 0 | 1 | ```\nclass TrieNode:\n def __init__(self):\n self.children = dict()\n self.end = False\n self.idx = -1\n self.palindromeIdxs = list()\n\nclass Solution:\n def __init__(self):\n self.root = TrieNode()\n \n def palindromePairs(self, words: List[str]) -> List[List[int]]:\... | 2 | You are given a **0-indexed** array of **unique** strings `words`.
A **palindrome pair** is a pair of integers `(i, j)` such that:
* `0 <= i, j < words.length`,
* `i != j`, and
* `words[i] + words[j]` (the concatenation of the two strings) is a palindrome.
Return _an array of all the **palindrome pairs** of_ `... | null |
Python solution using dictionary | palindrome-pairs | 0 | 1 | At first the solution seems to be pretty easy: we just check if reversed `words[i]` is in `words`. If yes and the index `j` of reversed word is not equal to `i`, correspondent indices `[i,j]` are added to the answer list. This covers the case like `"abcd"` and `"dcba"` , when both are in `words`\n\nIf index `j` of reve... | 2 | You are given a **0-indexed** array of **unique** strings `words`.
A **palindrome pair** is a pair of integers `(i, j)` such that:
* `0 <= i, j < words.length`,
* `i != j`, and
* `words[i] + words[j]` (the concatenation of the two strings) is a palindrome.
Return _an array of all the **palindrome pairs** of_ `... | null |
[Python] Trie solution explained | palindrome-pairs | 0 | 1 | ```\nclass TrieNode:\n def __init__(self):\n self.children = dict()\n self.end = False\n self.idx = -1\n self.palindromeIdxs = list()\n\nclass Solution:\n def __init__(self):\n self.root = TrieNode()\n \n def palindromePairs(self, words: List[str]) -> List[List[int]]:\... | 17 | You are given a **0-indexed** array of **unique** strings `words`.
A **palindrome pair** is a pair of integers `(i, j)` such that:
* `0 <= i, j < words.length`,
* `i != j`, and
* `words[i] + words[j]` (the concatenation of the two strings) is a palindrome.
Return _an array of all the **palindrome pairs** of_ `... | null |
336: Time 95.65%, Solution with step by step explanation | palindrome-pairs | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe algorithm uses a dictionary to store the reverse of the words and their indices, and then iterates through each word and checks for palindrome pairs using substrings. It first checks for empty string as a special case, a... | 8 | You are given a **0-indexed** array of **unique** strings `words`.
A **palindrome pair** is a pair of integers `(i, j)` such that:
* `0 <= i, j < words.length`,
* `i != j`, and
* `words[i] + words[j]` (the concatenation of the two strings) is a palindrome.
Return _an array of all the **palindrome pairs** of_ `... | null |
python solution better than 90 percent with comments using trie | palindrome-pairs | 0 | 1 | \tRuntime: 2768 ms, faster than 88.29% of Python3 online submissions for Palindrome Pairs.\n\tMemory Usage: 271.5 MB, less than 27.76% of Python3 online submissions for Palindrome Pairs.\n\t\n\t\n\tclass Solution:\n def palindromePairs(self, words: List[str]) -> List[List[int]]:\n \n trie = {}\n ... | 1 | You are given a **0-indexed** array of **unique** strings `words`.
A **palindrome pair** is a pair of integers `(i, j)` such that:
* `0 <= i, j < words.length`,
* `i != j`, and
* `words[i] + words[j]` (the concatenation of the two strings) is a palindrome.
Return _an array of all the **palindrome pairs** of_ `... | null |
recursive approach + memoization | house-robber-iii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:n\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:n\n<!-- Add your space complexity here, e.g. $$O(n)$$ ... | 6 | The thief has found himself a new place for his thievery again. There is only one entrance to this area, called `root`.
Besides the `root`, each house has one and only one parent house. After a tour, the smart thief realized that all houses in this place form a binary tree. It will automatically contact the police if ... | null |
[Python3] Dynamic Programming + Depth First Search | house-robber-iii | 0 | 1 | * we construct a dp tree, each node in dp tree represents [rob the current node how much you gain, skip the current node how much you gain]\n dp_node[0] =[rob the current node how much you gain]\n dp_node[1] =[skip the current node how much you gain]\n* we start the stolen from the leaf: Depth First Search\n* for each ... | 126 | The thief has found himself a new place for his thievery again. There is only one entrance to this area, called `root`.
Besides the `root`, each house has one and only one parent house. After a tour, the smart thief realized that all houses in this place form a binary tree. It will automatically contact the police if ... | null |
Recursive Logic Python3 5 Lines of Code | house-robber-iii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 8 | The thief has found himself a new place for his thievery again. There is only one entrance to this area, called `root`.
Besides the `root`, each house has one and only one parent house. After a tour, the smart thief realized that all houses in this place form a binary tree. It will automatically contact the police if ... | null |
Smallest Integer Divisible by K | house-robber-iii | 0 | 1 | Given a positive integer `K`, we must find the *length* of the smallest positive integer `N` such that `N` is divisible by `K`, and `N` only contains the digit `1`. If there is no such `N`, return `-1`. Therefore the following constraints must be met.\n\n* `N % K == 0`\n* length of `N` is minimized\n* `N` only contains... | 20 | The thief has found himself a new place for his thievery again. There is only one entrance to this area, called `root`.
Besides the `root`, each house has one and only one parent house. After a tour, the smart thief realized that all houses in this place form a binary tree. It will automatically contact the police if ... | null |
[Python3] Easy to understand | Recursive Post-order | Linear time & space | house-robber-iii | 0 | 1 | # Intuition\nAt each node, we either need to consider the node iteslf or its children.\n\n# Approach\nFrom each node, return the max we can get by including the node and also max without the node.\n\n1. For base cases (Leaf node\'s Null children), return `[0, 0]`\n2. With postorder traversal, at each node up in recusrs... | 2 | The thief has found himself a new place for his thievery again. There is only one entrance to this area, called `root`.
Besides the `root`, each house has one and only one parent house. After a tour, the smart thief realized that all houses in this place form a binary tree. It will automatically contact the police if ... | null |
EASY RECURSIVE PYTHON SOLUTION | house-robber-iii | 0 | 1 | # Approach\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can handle our problem with simple recursion. If we are at a node in our tree, we cannot rob the house if we robbed the house above it. (This creates our 2 cases) We can treat this recursive behavior by using the recursion fairy.\n\nThe... | 1 | The thief has found himself a new place for his thievery again. There is only one entrance to this area, called `root`.
Besides the `root`, each house has one and only one parent house. After a tour, the smart thief realized that all houses in this place form a binary tree. It will automatically contact the police if ... | null |
337: Solution with step by step explanation | house-robber-iii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe problem can be solved using a recursive approach. We can traverse the binary tree in a post-order traversal. At each node, we need to decide whether to rob the node or not.\n\nLet\'s define a helper function dfs which re... | 6 | The thief has found himself a new place for his thievery again. There is only one entrance to this area, called `root`.
Besides the `root`, each house has one and only one parent house. After a tour, the smart thief realized that all houses in this place form a binary tree. It will automatically contact the police if ... | null |
DP solution Python | house-robber-iii | 0 | 1 | ```\nfrom collections import defaultdict\n\nclass Solution:\n def rob(self, root: Optional[TreeNode]) -> int:\n def rec(node, depth):\n if not node:\n return\n layer_dict[depth].append(node)\n rec(node.left, depth + 1)\n rec(node.right, depth + 1)\n ... | 4 | The thief has found himself a new place for his thievery again. There is only one entrance to this area, called `root`.
Besides the `root`, each house has one and only one parent house. After a tour, the smart thief realized that all houses in this place form a binary tree. It will automatically contact the police if ... | null |
Dynamic Programming Solution | counting-bits | 0 | 1 | # Intuition\nKey intuition is the the value of the current number is dependent on previous numbers. Hence, we can divide this problem into sub-problems and we can use dynamic programming.\n\n# Approach\nWe can see if a number is even, it depends on the number x//2 before it, if it is uneven it depends on the number x//... | 1 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
✅ 97.97% DP + Bit Manipulation & Offset | counting-bits | 1 | 1 | # Interview Guide - Counting Bits in Binary Representation of Integers\n\n## Problem Understanding\n\n### Description\nThe problem at hand asks us to count the number of 1\'s in the binary representation of integers from 0 to `n` . It\'s a classic problem that combines elements of dynamic programming and bitwise opera... | 93 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
Python3 Solution | counting-bits | 0 | 1 | \n```\nclass Solution:\n def countBits(self, n: int) -> List[int]:\n dp=[0]\n for i in range(1,n+1):\n if i%2==1:\n dp.append(dp[i-1]+1)\n else:\n dp.append(dp[i//2])\n return dp \n``` | 5 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
✅97.54% Solution🔥Python3/C/C#/C++/Java🔥Use Vectors🔥 | counting-bits | 1 | 1 | # Code\n```Python3 []\nclass Solution:\n def countBits(self, n: int) -> List[int]:\n sum=0\n list1 = []\n for i in range(n+1):\n sum=bin(i).count("1")\n list1.append(sum)\n return list1\n```\n```python []\nclass Solution:\n def countBits(self, n: int) -> List[int]... | 16 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
Beginner-friendly || Simple solution with using Counting in Python3 | counting-bits | 0 | 1 | # Intuition\nThe problem description requires to convert the decimal into binary and represent the result as a **list of bits**.\n\nThough, the solution isn\'t **EFFICIENT** and there\'s a more affordable algorithm as **bit manipulation**, for the sake of brevity, we\'ll skip it and will focus on counting.\n\n---\n\nIf... | 1 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
【Video】Time: O(n), Space: O(n) solution with Python, JavaScript, Java and C++ | counting-bits | 1 | 1 | # Intuition\nThe most important idea to solve the question of counting the number of set bits (1s) in the binary representation of numbers from 0 to n is to recognize and utilize the pattern that relates the count of set bits in a number to the count of set bits in another number that is a power of 2. This concept allo... | 33 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
easy C++/Python bit Manipulations||welcome for tricks | counting-bits | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nProvide $O(n)$ linear time solutions. __builtin_popcount or C++ bitset count() are supposed $O(\\log n)$ time, so for fast implementation are not used.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n[Please turn on ... | 7 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
[Python] Simple Solution with Clear Explanation | counting-bits | 0 | 1 | # Solution Code\n```\ndef countBits(self, num: int) -> List[int]:\n counter = [0]\n for i in range(1, num+1):\n counter.append(counter[i >> 1] + i % 2)\n return counter\n```\n\n# Analysis\nTo understand the solution, we remember the following two points from math:\n* All whole numbers can be represented... | 495 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
Python | using Bin() | Easy to understand | counting-bits | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def countBits(self, n: int) -> List[int]:\n ans = [0] * (n+1) // create a list with n+1 of length\n count = 0 \n for i in range(1,n+1): //note: binary of 1 will always be 0 so start from position 1 to n+1\n binary = bin(i)[2:] // convert number into ... | 1 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
Python | Easy to Understand | 4 Lines of Code | counting-bits | 0 | 1 | # Python | Easy to Understand | 4 Lines of Code\n```\nclass Solution:\n def countBits(self, n: int) -> List[int]:\n result=[0]*(n+1) \n offset=1\n for i in range(1,n+1):\n if offset*2 ==i:\n offset=i\n result[i]=1+result[i-offset]\n \n retu... | 2 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
Beginner-friendly || Two solutions || Counting || Dynamic Programming || Python3 & TypeScript | counting-bits | 0 | 1 | # Intuition\nThe problem description requires to convert the decimal into binary and represent the result as a **list of bits**.\n\n---\n\nIf you haven\'t familiar with [binary](https://en.wikipedia.org/wiki/Binary_number), follow the link.\n\n---\n\nWe\'d like to consider two solutions (**Dynamic Programming** and **C... | 1 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
Python 1-liner. Functional programming. | counting-bits | 0 | 1 | # Approach\nNotice, that the number of `1s` in a number `n` is equal to number of `1s` in `n // 2`, if `n` is even else one more than it.\n\ni.e $$bits(n) = bits(n \\div 2) + (n \\bmod 2)$$\n\n# Complexity\n- Time complexity: $$O(n)$$\n\n- Space complexity: $$O(n)$$\n\n# Code\nImperative multi-liner:\n```python\nclass ... | 1 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
Simple straight forward solution in Python3 | counting-bits | 0 | 1 | # Intuition\nThe array in range of (2^i)+1 to 2^(i+1) can be gernerate from previous range from (0, 2^i).\n\n# Code\n```\nclass Solution:\n def countBits(self, n: int) -> List[int]:\n res = [0]\n\n i = 0\n while(2**i <= n):\n for j in range(2**i):\n res.append(res[j]+1)... | 1 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
Python3 | Very simple solution | For beginners | counting-bits | 0 | 1 | # Intuition\nProblem solution used ***"191. Number of 1 Bits"*** \nand a shell is written with complexity ***[O(n) * O(n)]***\n\n# Approach\n***We feed each number of the function sequence and form a new list from this, which we then return - everything is very simple***\n\n# Complexity\n- Time complexity:\n$$O(n^2)$$\... | 1 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
O(n) USING MAP() IN PYTHON | counting-bits | 0 | 1 | # Intuition:\nThe problem involves counting the number of set bits (1s) in the binary representation of numbers from 0 to n.\n\n# Approach:\nThe provided solution employs a for loop to iterate through the range from 0 to n (inclusive). For each number, it converts it to binary using bin(i)[2:] to remove the \'0b\' pref... | 1 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
Counting Bits | Bit manipulation | python3 | | counting-bits | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
🔥🔥🔥Python Easy Brute Force🔥🔥🔥 | counting-bits | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 8 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
Superb Logic Solution | counting-bits | 0 | 1 | \n\n# Solution In Python3\n```\nclass Solution:\n def countBits(self, n: int) -> List[int]:\n sum=0\n list1 = []\n for i in range(n+1):\n sum=bin(i).count("1")\n list1.append(sum)\n return list1\n```\n# Please upvote me it would encourages me | 3 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
Beats 100% | Easy solution using bit manipulation | counting-bits | 1 | 1 | ```C++ []\nclass Solution {\npublic:\n vector<int> countBits(int n) {\n vector<int> ans(n + 1, 0);\n \n for (int i = 1; i <= n; i++) {\n // To count the number of 1\'s in the binary representation of i,\n // you can use the fact that ans[i] = ans[i / 2] + (i % 2).\n ... | 4 | Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
**Example 1:**
**Input:** n = 2
**Output:** \[0,1,1\]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
**Example 2:**
**Input:** n =... | You should make use of what you have produced already. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous. Or does the odd/even status of the number help you in calculating the number of 1s? |
✅ 95.83% Recursive Flattening & Stack | flatten-nested-list-iterator | 1 | 1 | # Intuition\nWhen we hear "nested structure", the brain immediately thinks of recursion or stacks. Why? Because both tools excel at handling layered structures. Imagine Russian nesting dolls. To access the innermost doll, we must open each outer doll. This problem gives us a nested list and asks us to flatten it. A nes... | 93 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
🚀 94.83% || Recursion || Explained Intuition 🚀 | flatten-nested-list-iterator | 1 | 1 | # Probelm Description\n\nThe task is designing a class called `NestedIterator` for flattening a nested list of integers. The nested list contains elements, where each element can either be an integer or another nested list with its own elements. The goal is to create an iterator that can traverse this nested structure ... | 52 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
✅☑[C++/Java/Python/JavaScript] || Beats 100% || EXPLAINED🔥 | flatten-nested-list-iterator | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n1. **Private Member Variables:**\n\n - **nestedList:** A vector of `NestedInteger` objects that represents the input nested list.\n - **tempList:** A vector of integers used to store the flattened list temporarily.\n - **I... | 3 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
Python3 Solution | flatten-nested-list-iterator | 0 | 1 | \n```\n# """\n# This is the interface that allows for creating nested lists.\n# You should not implement it, or speculate about its implementation\n# """\n#class NestedInteger:\n# def isInteger(self) -> bool:\n# """\n# @return True if this NestedInteger holds a single integer, rather than a nested list... | 3 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
Video Solution | Explanation With Drawings | In Depth | Java | C++ | Python 3 | flatten-nested-list-iterator | 1 | 1 | # Intuition, approach, and complexity discussed in detail in video solution\nhttps://youtu.be/2zpPPM6W7IA\n\n# Code\n\nC++\n```\nclass NestedIterator {\n vector<int> flatList;\n int iterator;\npublic:\n NestedIterator(std::vector<NestedInteger>& nestedList) {\n iterator = 0;\n flattenList(nestedL... | 2 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
Recursion, Class pointer to itself | C++, Python | flatten-nested-list-iterator | 0 | 1 | My idea was to use an index to point to the elements in the `nestedList` that\'s passed in, if the current element `idx` points to holds a single integer, than we could just return that.\nBut what are we going to do if the `NestedInteger` that I\'m currently point to does not contain a single integer but holds another ... | 1 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
simple python intutive solution | flatten-nested-list-iterator | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
Simple Python Solution Using `yield` with O(1) space | flatten-nested-list-iterator | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nPython has built-in iterator support, why not use it?\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nDefine a Python Iterator using yield. This solution can implicitly handles empty lists. The only issue is that P... | 1 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
Easiest solution in Python | flatten-nested-list-iterator | 0 | 1 | # Intuition\nThere is no real intuition simply following the instructions.\n\n# Code\n```\n# """\n# This is the interface that allows for creating nested lists.\n# You should not implement it, or speculate about its implementation\n# """\n#class NestedInteger:\n# def isInteger(self) -> bool:\n# """\n# ... | 1 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
Easy to Understand || c++ || java || python | flatten-nested-list-iterator | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nHere\'s a step-by-step explanation of the C++ code\'s approach:\n\nThe NestedIterator constructor takes a vector of NestedInteger as input.\n\nIn the constructor, it pushes all elements of the input list onto the stack in re... | 29 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
【Video】Give me 10 minutes - How we think about a solution - Python, Javascript, Java, C++ | flatten-nested-list-iterator | 1 | 1 | # Intuition\nUsing stack\n\n---\n\n# Solution Video\n\nhttps://youtu.be/N8QehVXYSc0\n\n\u25A0 Timeline of the video\n`0:04` 2 Keys to solve this question\n`0:18` Explain the first key point\n`0:45` Explain the first key point\n`1:27` Explain the second key point\n`5:44` Coding\n`9:39` Time Complexity and Space Complexi... | 16 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
Simple to understand Python Solution | flatten-nested-list-iterator | 0 | 1 | # Intuition\nRecursion(DFS) and keeping track of index in flattened list\n\n# Approach\n**1) Flatten part**\n->lets say we have flattened_list = []\nyou want to flatten [4,[1,2]]\nwe can see that 4 is not a list but an integer so we can simply\nappend it into flattened_list\n\n->Now we are left with [1,2] but [1,2] its... | 0 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
Lazy Recursive Solution w/ Flat Generator | flatten-nested-list-iterator | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUse generators to perform lazy flattening of a large and deep `nestedList`. \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUse `yield` to produce integers and `yield from` to produce integers from a nested genera... | 0 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
Cheat solution with Generators in Python | flatten-nested-list-iterator | 0 | 1 | # Intuition\nGenerators in Python itself works very similar to iterators, so one can just reuse it. In case of flatten list - just yield the elements from it, in case of nested structure - call the generator recursively.\nTo support hasNext method, one can take the next element in advance, to check if it exists.\n\n# C... | 1 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
341: Time 90.75%, Solution with step by step explanation | flatten-nested-list-iterator | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe NestedIterator class implements an iterator to flatten a nested list of integers. It has three methods:\n\n1. __init__(self, nestedList: List[NestedInteger]): Initializes the iterator with the nested list nestedList. It ... | 8 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
easy C++/Python recursion based on array vs queue|| 3ms Beats 94.83% | flatten-nested-list-iterator | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUse recursion. Divide & conquer!\nFirst appeoach uses vector, second C++ solution uses queue.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nLet\'s see how the recursion flatten function `f` works. Consider the nest... | 7 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
[Python] Generator Solution with Explanation | flatten-nested-list-iterator | 0 | 1 | ### Introduction\n\nGiven an unknown implementation of a nested list structure containing integer values, design an iterator as a wrapper for the nested list structure which allows for 1) checking if a next value exists, and 2) obtaining of the next value in the iteration.\n\nI\'d just like to note that the type hintin... | 52 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
Python 3 stack with memory-efficient iterator and lazy evaluation | flatten-nested-list-iterator | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nWe can use a stack to manage iterators for each level of nested lists. The `hasNext` method looks at the top iterator on the stack and tries to fetch the next element. If it\'s an integer, `hasNext` stores it for the next method to retu... | 3 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
🔥 Optimizing Nested List Iteration: Three Approaches for Efficient Flattening || Mr. Robot | flatten-nested-list-iterator | 1 | 1 | \n# Flattening Nested Lists: Exploring the "NestedIterator" Class \u2753\n---\n## \uD83D\uDCA1Approach 1: Using Stack\n\n### \u2728Explanation\n\nThe `NestedIterator` class is a handy utility for dealing with nested lists in C++. This blog post dives into how this class can be used to efficiently flatten nested lists, ... | 1 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
Python short and clean. Functional programming. | flatten-nested-list-iterator | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n\n- Space complexity: $$O(n)$$\n\nwhere,\n`n is number of integers in the nested_list`.\n\n# Code\n```python\nclass NestedIterator:\n def __init__(self, nested_list: list[NestedInteger]):\n self.xs = flatten(nested_list)\n self.next_x = None\n \n def nex... | 1 | You are given a nested list of integers `nestedList`. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.
Implement the `NestedIterator` class:
* `NestedIterator(List nestedList)` Initializes the iterator with the nested list `nestedLi... | null |
【Video】Give me 3 minutes - without bitwise operations - Python, JavaScript, Java, C++ | power-of-four | 1 | 1 | # Intuition\nUse math.log function\n\n---\n\n# Solution Video\n\nhttps://youtu.be/dy_GA7sG22g\n\n\u25A0 Timeline of the video\n`0:05` Easy way to solve a constraint\n`0:03` Example code with Bitwise Operations\n`0:19` Solution with Build-in function\n`1:23` Coding with Build-in function\n`1:50` Time Complexity and Spac... | 36 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
Python3 Solution | power-of-four | 0 | 1 | \n```\nclass Solution:\n def isPowerOfFour(self, n: int) -> bool:\n if n<=0:\n return False\n return n>0 and log(n,4).is_integer() \n``` | 3 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
Video Solution | Explaination With Drawings | In Depth | Java | C++ | Python 3 | power-of-four | 1 | 1 | # Intuition, approach, and complexity discussed in detail in video solution.\nhttps://youtu.be/GWcEoppPzBQ\n\n# Code\nC++\n```\n//TC : O(1)\n//SC : O(1)\nclass Solution {\npublic:\n bool isPowerOfFour(int n) {\n return n > 0 && floor(log(n)/log(4)) == ceil(log(n)/log(4));\n }\n};\n```\nJava\n```\nclass So... | 3 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
🚀 100% || Brute Force & Two Math Solutions || Explained Intuition 🚀 | power-of-four | 1 | 1 | # Problem Description\n\nGiven an integer `n`. The task is to determine whether `n` is a power of **four**. \nAn integer `n` is considered a power of **four** if there exists another integer `x` such that `n` is equal to `4` raised to the power of `x` (i.e., $n = 4^x$).\nReturn `true` if `n` is a power of **four**; oth... | 107 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
🔥Shortest, Easiest 1 Line Solution | power-of-four | 0 | 1 | # Approach\nWe have to verify if an integer is a power of 4 by checking if it\'s both positive and if its logarithm with base 4 is an integer; if so, it\'s considered a power of 4.\n# Code\n```py\nclass Solution:\n def isPowerOfFour(self, n: int) -> bool:\n return n>0 and log(n,4).is_integer()\n```\n# Line by... | 5 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
✅☑[C++/Java/Python/JavaScript] || Beats 100% || 1 line Code || 3 Approaches || EXPLAINED🔥 | power-of-four | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1 (Brute force)***\n1. The `isPowerOfFour` function takes an integer `n` as input and returns a boolean (true or false) based on whether n is a power of four or not.\n\n1. It initializes a `for` loop that iterat... | 2 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
Bitwise Solution | Python | power-of-four | 0 | 1 | # Code\n```\nclass Solution:\n def isPowerOfFour(self, n: int) -> bool:\n return n.bit_count() == 1 and int(\'10\' * 16, 2) & n == 0\n``` | 1 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
Python Solution Using Logarithms | power-of-four | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe are dealing with powers of 4. We know that...\n4 ^ 0 = 1, \n4 ^ 1 = 4, \n4 ^ 2 = 16, etc.\nIn math, we use logarithms to find how many times the base number (in our case, 4) must be multiplied by itself to get some other particular num... | 1 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
Beginner-friendly || Loops and Math || Simple solution with Python3 / TypeScript | power-of-four | 0 | 1 | # Intuition\nLet\'s briefly explain what the problem is:\n- there\'s an integer `n`\n- our goal is to define, if this integer is a possible answer to `n = 4^x`\n\nThere\'re some solutions, including math and loops.\n\n# Complexity\n- Time complexity: **O(log N)** with loop solution, and **O(1)** - with logarithmic solu... | 1 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
✅ One liner✅Beats 100%✅ easy and simple solution with explanation | power-of-four | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis code checks if a given number n is a power of four. In other words, it determines if n can be expressed as 4 raised to some non-negative integer (e.g., 1, 4, 16, 64, etc.).\n# Approach\n<!-- Describe your approach to solving the prob... | 15 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
✅ 6 Different Methods || Beats 99.76% 🔥 || | power-of-four | 1 | 1 | ## This Solution took me a lot of time, so please UpVote\u2705 if you found this helpful :)\n# Approach 1: Iteration\n\n\n# Intuition : \n**The idea is to keep dividing the... | 36 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
Python 🐢 Slow and Fast 🐇 1 Lines without loops | power-of-four | 0 | 1 | # Target\nTo determine if an integer is a power of four, we examine its binary representation.\n\n# Approach\n- First, we ensure the input integer n is positive; negative numbers cannot be powers of four.\n- We check if there is only one set bit in n by verifying that n & (n - 1) equals zero. This condition ensures a s... | 0 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
One liner Bitwise Manipulation | power-of-four | 0 | 1 | # Intuition\nThe problem requires us to check if a given integer is a power of 4. To understand whether a number is a power of 4, we can make use of bitwise operations. We know that a power of 4 is a positive integer with a binary representation that has only one set bit, and that set bit is at an odd position.\n# Appr... | 2 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
Easy. One Line Python Solution. Logarithm. | power-of-four | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUse math\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUse n logarithm to base 4.\n1. If n is negative return False.\n2. Check if the n logarithm to base 4 belongs to naturale numbers or not.\n$$log4(n)\u2208N$$\n#... | 0 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
1-liner | power-of-four | 0 | 1 | \n\n# Code\n```\ndef isPowerOfFour(self, n: int) -> bool:\n return n==1 if n<=1 else int(sqrt(n)) * int(sqrt(n)) == n and int(sqrt(n)).bit_count() == 1\n``` | 0 | Given an integer `n`, return _`true` if it is a power of four. Otherwise, return `false`_.
An integer `n` is a power of four, if there exists an integer `x` such that `n == 4x`.
**Example 1:**
**Input:** n = 16
**Output:** true
**Example 2:**
**Input:** n = 5
**Output:** false
**Example 3:**
**Input:** n = 1
**O... | null |
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