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
values |
|---|---|---|---|---|---|---|---|
✅ 99.53% Recursion & Math & Dynamic Programming | pascals-triangle | 1 | 1 | # Comprehensive Guide to Generating Pascal\'s Triangle: A Three-Pronged Approach for Programmers\n\n## Introduction & Problem Statement\n\nWelcome to this in-depth guide on generating Pascal\'s Triangle up to a given number of rows. Pascal\'s Triangle is a mathematical concept that finds applications in various domains... | 74 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
Python3 Code: | pascals-triangle | 0 | 1 | \n# Code\n```\nclass Solution:\n def generate(self, numRows: int) -> List[List[int]]:\n triangle = [[1]] #creating a new row, as it always begins with one, initialising that alone\n for i in range(1,numRows): #for loop for the range of 1 to the num of rows given\n prev_row = triangle[-1] #as... | 1 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
Brute-force Approach (Runtime - beats 93.54%) | Simple Straightforward Solution | pascals-triangle | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code aims to generate Pascal\'s Triangle up to the specified number of rows. Pascal\'s Triangle is a triangular array of numbers in which each number is the sum of the two numbers directly above it. The first and last values in each r... | 1 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
Beginner Friendly Python Solution | pascals-triangle | 0 | 1 | # Code\n```\nclass Solution:\n def generate(self, numRows: int) -> List[List[int]]:\n triangle = []\n for i in range(numRows):\n row = [1] * (i + 1)\n for j in range(1, i):\n row[j] = triangle[i - 1][j - 1] + triangle[i - 1][j]\n triangle.append(row)\n ... | 1 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
[ Java / Python / C++] 🔥100% | ✅EASY | pascals-triangle | 1 | 1 | \n``` java []\nclass Solution {\n public List<List<Integer>> generate(int n) {\n List<List<Integer>> dp = new ArrayList<>();\n\n for (int i = 0; i < n; ++i) {\n Integer[] temp = new Integer[i + 1];\n Arrays.fill(temp, 1);\n dp.add(Arrays.asList(temp));\n }\n\n for (int i = 2; i < n; ++... | 2 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
Generating Pascal's Triangle in C++, Python, and Dart 😃 | pascals-triangle | 0 | 1 | # Intuition\nMy initial thoughts on solving this problem revolve around traversing through the array and keeping track of the maximum sum obtained so far. I might consider using Kadane\'s algorithm, which efficiently finds the maximum subarray sum in a single traversal.\n\n# Approach\n1. Initialize variables: max_sum t... | 6 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
【Video】Beats 97.21% - Python, JavaScript, Java, C++ | pascals-triangle | 1 | 1 | # Main point of the Solution\n\nThe main point of this solution is to generate Pascal\'s triangle, up to a specified number of rows, by iteratively constructing each row based on the previous one. It initializes with the first row containing a single "1," then iterates through the desired number of rows, creating each ... | 24 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
Python | Easy to Understand | Fast | pascals-triangle | 0 | 1 | # Python | Easy to Understand | Fast\n```\nclass Solution:\n def generate(self, numRows: int) -> List[List[int]]:\n finalNums=[]\n finalNums.append([1])\n for i in range(numRows-1):\n newRow=[1]\n for j in range(i):\n newRow.append(finalNums[i][j]+finalNums[i... | 17 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
🚀 C++beats 100% 0ms || Java || Python || DP & Combinatorics || Commented Code 🚀 | pascals-triangle | 1 | 1 | # Problem Description\nGiven an integer numRows, return the first `numRows` of **Pascal\'s triangle**.\n\nIn **Pascal\'s triangle**, each number is the sum of the two numbers directly above it as shown:\n\n.\n\n# Strategies to Tackle the Problem:\n1. **Understand the Goal**: First, understand that the code\'s goal is to generate Pascal\'s Triangle up to a specified number of rows (`numRows`).\n\n1. **Visualize Pasc... | 7 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
Simple recursion Python | pascals-triangle | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def generate(self, numRows: int) -> List[List[int]]:\n if numRows == 1:\n return [[1]]\n prev = self.generate(numRows - 1)\n fin = prev[-1]\n now = [1]\n for i in range(len(fin)-1):\n now.append(fin[i] + fin[i+1])\n no... | 4 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
Easy || 0 ms || 100% || Fully Explained || Java, C++, Python, JavaScript, Python3 || DP | pascals-triangle | 1 | 1 | # **Java Solution:**\nRuntime: 0 ms, faster than 100.00% of Java online submissions for Pascal\'s Triangle.\n```\nclass Solution {\n public List<List<Integer>> generate(int numRows) {\n // Create an array list to store the output result...\n List<List<Integer>> output = new ArrayList<List<Integer>>();\... | 217 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
Bottom-up tabulation (Dynamic Programming) in Python3 | pascals-triangle | 0 | 1 | # Intuition\nThe problem description is the following:\n- we have an integer `numRows`\n- our goal is to recreate the **Pascal\'s triangle (PT)**\n\n\n---\n\nIf you want to know more about [PT](https://en.wikipedia.org/wiki/Pascal%27s_triangle), follow the link.\n\n```\n# Example\nnumRows = 4\n\n# The first level is al... | 1 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
Beats 100%||C/C++/Python iterative DP||Math Explain | pascals-triangle | 0 | 1 | # Intuition\nHere is a real C solution provided which is really very rare.\n\nSolution via Pascal\'s identity $$C^i_j=C^{i-1}_{j-1}+C^{i-1}_{j}$$& $$C^i_j=C^i_{i-j}$$\n\nDon\'t use the built-in function comb(i, j) to compute, because it is very time consuming.\n\nIt is also not suggested direct to use the defintion to ... | 10 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
Pascals Traingle (simple Maths) | pascals-triangle | 0 | 1 | # Intuition\n<!-- Describe your first- thoughts on how to solve this problem. -->\nUsing the for loops and just iterating through it..\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\njust observe the pattern and see for every row the leading and ending 1\'s are in commmon..\n\n# Complexity\n- Ti... | 5 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
👹 easy simple solution, python & java 🐍☕ | pascals-triangle | 1 | 1 | # Intuition\nstarts with 1, then goes down a level and sums the two values and puts in after the first one, and incremends for each row\n\n# Approach\nfirst row is always 1\ngo row by columns, ie: nested for loop\n1. for i in range (from 1 to rowsnumber):\'\n inside the loop add a 1 as the first elem in the first co... | 3 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
Easy Python Solution With Briefly Explanation ✅✅ | pascals-triangle | 0 | 1 | # Approach\nLet\'s break down the given Python code step by step. This code defines a class `Solution` with a method `generate` that generates Pascal\'s triangle up to a specified number of rows.\n\n```python\nclass Solution:\n def generate(self, numRows: int) -> List[List[int]]:\n```\n\nThis line defines a class ca... | 5 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
NOOB CODE | pascals-triangle | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Initializes lists a1, a2, and an with the first three rows of Pascal\'s Triangle.\n2. Sets p to an to track the previous row.\n3. Creates an empty list f to store the entire Pascal\'s Triangle.\n4. Appends the first two r... | 3 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
Pascals Traingle (simple Maths) | pascals-triangle | 0 | 1 | # Intuition\n<!-- Describe your first- thoughts on how to solve this problem. -->\nUsing the for loops and just iterating through it..\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\njust observe the pattern and see for every row the leading and ending 1\'s are in commmon..\n\n# Complexity\n- Ti... | 1 | Given an integer `numRows`, return the first numRows of **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** numRows = 5
**Output:** \[\[1\],\[1,1\],\[1,2,1\],\[1,3,3,1\],\[1,4,6,4,1\]\]
**Example 2:**
**Input:** numRows = ... | null |
Python3 Solution | pascals-triangle-ii | 0 | 1 | \n```\nclass Solution:\n def getRow(self, rowIndex: int) -> List[int]:\n ans=[1]\n for i in range(1,rowIndex+1):\n ans.append(1)\n for j in range(len(ans)-2,0,-1):\n ans[j]+=ans[j-1]\n\n return ans\n``` | 3 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
【Video】Give me 10 minutes - How we think about a solution - Python JavaScript, Java, C++ | pascals-triangle-ii | 1 | 1 | Welcome to my post! This post starts with "How we think about a solution". In other words, that is my thought process to solve the question. This post explains how I get to my solution instead of just posting solution codes or out of blue algorithms. I hope it is helpful for someone.\n\n# Intuition\nadd left number and... | 40 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
✅ 100% Easy Optimized | pascals-triangle-ii | 1 | 1 | # Intuition\nWhen faced with Pascal\'s Triangle, a fundamental pattern that stands out is the binomial coefficient. The elements of each row in Pascal\'s Triangle can be represented as combinations. Recognizing this mathematical property offers a direct route to the solution without the need to iterate through all the ... | 72 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
📢-: The Magic of Pascal's Triangle: Generating Rows with Ease || Mr. Robot | pascals-triangle-ii | 1 | 1 | # Efficiently Generating Pascal\'s Triangle Row\n\n## Introduction\n\nIn the world of combinatorics, Pascal\'s Triangle is a true gem. It\'s a triangular array of numbers where each number is the sum of the two numbers directly above it. In this blog post, we\'ll explore an efficient C++ code that generates the `rowInd... | 1 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
Video Solution | Explanation With Drawings | In Depth | C++ | Java | Python 3 | pascals-triangle-ii | 1 | 1 | # Intuition and approach discussed in detail in video solution\nhttps://youtu.be/r-7NUInsUjA\n# Code\nC++\n```\nclass Solution {\npublic:\n vector<int> getRow(int rowIndex) {\n vector<vector<int>> tria (rowIndex + 1);\n int colNum = 1;\n for(int r = 0; r <= rowIndex; r++){\n \n ... | 1 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
Pascal’s Triangle Solution in python with explanation | pascals-triangle-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n##### Points to remember after reading the problem:\n1. Every row depends on previous row.\n2. First two rows are [1] and [1,1]\n3. They asked only the particular row to return, so we will go through iteration for every row until given in... | 1 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
Easy Python3 solution | pascals-triangle-ii | 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. -->\nwe know that first and last value in any row is 1, then we can build\nthe tree till the rowIndex by iterating over previous res row and\nappend to new row created ever... | 1 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
This is the Best Approch To Find The Solution In the Best Time Complexity..... | pascals-triangle-ii | 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 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
4 line using math | pascals-triangle-ii | 0 | 1 | # Formula\n$$ans[0] = 1$$\n$$k = 1 \\,\\to\\, n$$\n$$ans[k]=ans[k-1]\xD7 \\frac{n\u2212k+1}k\\\n\n\u200B\t\n $$\n# Code\n```py\nclass Solution:\n def getRow(self, n: int) -> List[int]:\n ans = [1]\n for k in range(1, n + 1):\n ans.append(ans[k-1] * (n - k + 1) // k)\n return ans\n```\... | 1 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
Beginner-friendly || Simple solution with Dynamic Programming in Python3 / TypeScript | pascals-triangle-ii | 0 | 1 | # Intuition\nLet\'s briefly explain what the problem is:\n- there\'s a `rowIndex` of **Pascal\'s Triangle**\n- our goal is to extract a particular row by `rowIndex`\n\nTo be short, [Pascal\'s Triangle](https://en.wikipedia.org/wiki/Pascal%27s_triangle) is a matrix of rows, whose elements follow the next formula.\n\n```... | 1 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
Quick & Easy Python Solution | pascals-triangle-ii | 0 | 1 | \n# Python solution\n```\nclass Solution:\n def getRow(self, rowIndex: int) -> List[int]:\n results=[1]\n last=1\n for i in range(1,rowIndex+1):\n next_value=last * (rowIndex - i + 1) // i\n results.append(next_value)\n last=next_value\n return results\n``... | 1 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
✅ Beats 98.39% 🔥 within 4 Lines || Easiest Solution 2 Approaches || | pascals-triangle-ii | 1 | 1 | \n\n\n# Intuition\n1. We know that the pascal triangle is a triangle of binomial coefficients.\n2. We know that the binomial coefficients can be calculated using the formula nCr = n!/(r!*(n-r)!)\n3. We know ... | 1 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
🚀 100 % || DP & Combinatorics then Optimized Space || Commented Code 🚀 | pascals-triangle-ii | 1 | 1 | # Problem Description\nGiven an integer rowIndex, return the `rowIndexth` (0-indexed) row of the **Pascal\'s triangle**.\n\nIn **Pascal\'s triangle**, each number is the sum of the two numbers directly above it as shown:\n\n row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
Beats 100% | Optimised Solution Using Combinations | pascals-triangle-ii | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIn this solution, we use the concept of combinations (n choose k) to calculate each element of the Pascal\'s Triangle\'s row. We start with a value of 1 and update it in each step by multiplying by the relevant factor to get the next valu... | 6 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
🔥5 - Approaches 🚀| Java | C++ | Python | JavaScript | C# | | pascals-triangle-ii | 1 | 1 | Here\'s the problem statement:\n\nGiven an integer `k`, return the `k`-th row of the Pascal\'s triangle.\n\nNow, let\'s discuss the approaches to solve this problem from naive to efficient:\n\n## 1. Naive Approach: 79% Beats\uD83D\uDE80\n - Generate Pascal\'s triangle up to the `k`-th row and return the `k`-th row.\n... | 33 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
Solution of pascal's triangle II problem | pascals-triangle-ii | 0 | 1 | # Approach\nSolved using dynamic programming\n- Step one: append 1 to list\n- Step two: from the end of the row construct a row for Pascal\'s triangle \n- Step three: continue until `len(triangle) != idx + 1`\n# Complexity\n- Time complexity:\n$$O(rowIndex)$$ - as we use extra space for answer equal to a constant\n\n- ... | 1 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
100% Acceptance | Optimised Solution Using a single array with reduced calculations | pascals-triangle-ii | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIt aims to optimize the generation of the `rowIndex`-th row of Pascal\'s Triangle by using a single array and a middle-out approach. By calculating elements from the center outwards, it reduces the number of calculations and minimizes mem... | 6 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
✔️ Python Solution with Briefly Explanation 🔥🔥 | pascals-triangle-ii | 0 | 1 | # Explanation\nLet\'s break down the given code step by step.\n\n```python\ndef getRow(self, rowIndex: int) -> List[int]:\n res = [[1]]\n```\n- This function is defined as a method within a class, as it has `self` as its first parameter. It takes an integer `rowIndex` as input and is expected to return a list of int... | 5 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
three liner || code || python || super easy approach | pascals-triangle-ii | 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. $$O(n)$$ --... | 0 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
🚀🚩Beats 100%🧭 | Easy and Optimised Solution🚀 Using a single array with less divisions🔥 | pascals-triangle-ii | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIt is an optimized approach to generate the `rowIndex`-th row of Pascal\'s Triangle. It utilizes a single array to store the values of each element in the row while minimizing calculations and avoiding overflow by using `long long` data t... | 5 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
The Finest solution | pascals-triangle-ii | 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 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
Python 99.1% beats 7lines code || Simple and easy approach | pascals-triangle-ii | 0 | 1 | **If you got help from this,... Plz Upvote .. it encourage me**\n# Code\n```\nclass Solution:\n def getRow(self, rowIndex: int) -> List[int]:\n prev = []\n for i in range(rowIndex+1):\n temp = [1]*(i+1)\n for j in range(1,i):\n temp[j] = prev[j] + prev[j-1]\n prev ... | 3 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
🚀📈Beats 100%🧭🚩 | 🔥Easy O(rowIndex) Solution 🚀| Optimised Solution 🚩 | pascals-triangle-ii | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this optimized code is to calculate the `rowIndex-th` row of Pascal\'s Triangle efficiently. Instead of constructing the entire triangle, we can iteratively calculate each element of the row in place.\n# Approach\n<!-... | 4 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
Python Easy Solution || Beginner Friendly | pascals-triangle-ii | 0 | 1 | # Code\n```\nclass Solution:\n def pascal(self, lis: list,k: int)-> List:\n res=[0]*(k+1)\n res[0]=1\n res[k]=1\n for i in range(1,len(res)-1):\n res[i]=lis[i-1]+lis[i]\n return res\n\n def getRow(self, rowIndex: int) -> List[int]:\n prev=[1]\n for i in ... | 1 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
✅ Beats 98.39% 🔥 | 2 Approaches Detailed Explanation | Beginner Friendly Code | | pascals-triangle-ii | 1 | 1 | \n\n\n# Intuition\n1. We know that the pascal triangle is a triangle of binomial coefficients.\n2. We know that the binomial coefficients can be calculated using the formula nCr = n!/(r!*(n-r)!)\n3. We know ... | 3 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
Approach made Easier ! | pascals-triangle-ii | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# ****Read the Approach : Looks long But it\'s easy ! :****\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\nAs you see :\n```***********\n\n# Step 1 :\n\nfor row 0 : we have 1 elements ;\n\nand for row 1 : we h... | 4 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
Normal Math Approach Two Approches --->Python3 | pascals-triangle-ii | 0 | 1 | \n# Brute Force\n```\nclass Solution:\n def getRow(self, rowIndex: int) -> List[int]:\n list1=[]\n for i in range(rowIndex+1):\n level=[]\n for j in range(i+1):\n if j==0 or j==i:\n level.append(1)\n else:\n level... | 3 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
My first solution in Python. | pascals-triangle-ii | 0 | 1 | Here is my successfully uploaded solution of today\'s problem in Python language.\n\n# Code\n```\nclass Solution:\n def getRow(self, rowIndex: int) -> List[int]:\n def insum(L):\n l=[1]\n for i in range(0,len(L)-1):\n l.append(L[i]+L[i+1])\n l.append(1)\n ... | 2 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
8 lines dp solution python3 | pascals-triangle-ii | 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. $$O(n)$$ --... | 7 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
Python easy recursion | 9 Line | 40 ms | pascals-triangle-ii | 0 | 1 | # Intuition\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n$$O(k^2)$$ where k is lengt of array with result\n- Space complexity:\n$$O(k)$$ \n\n# Code\n```\nclass Solution:\n def getRow(self, rowIndex: int) -> List[int]:\n if rowIndex == 0:\n ... | 1 | Given an integer `rowIndex`, return the `rowIndexth` (**0-indexed**) row of the **Pascal's triangle**.
In **Pascal's triangle**, each number is the sum of the two numbers directly above it as shown:
**Example 1:**
**Input:** rowIndex = 3
**Output:** \[1,3,3,1\]
**Example 2:**
**Input:** rowIndex = 0
**Output:** \[... | null |
Triangle-(python easy to understand) | triangle | 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. $$O(n)$$ -->\nO(n^2)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O... | 1 | Given a `triangle` array, return _the minimum path sum from top to bottom_.
For each step, you may move to an adjacent number of the row below. More formally, if you are on index `i` on the current row, you may move to either index `i` or index `i + 1` on the next row.
**Example 1:**
**Input:** triangle = \[\[2\],\[... | null |
Python 3 Solutions | triangle | 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 | Given a `triangle` array, return _the minimum path sum from top to bottom_.
For each step, you may move to an adjacent number of the row below. More formally, if you are on index `i` on the current row, you may move to either index `i` or index `i + 1` on the next row.
**Example 1:**
**Input:** triangle = \[\[2\],\[... | null |
[Python] In-place DP with Explanation | triangle | 0 | 1 | ### Intuition\n\nWe can use the ```triangle``` array to perform our DP. This allows us to use O(1) auxiliary space. Why?\n\n- For any given coordinate in the triangle ```(x, y)``` where ```0 <= x < len(triangle)``` and ```0 <= y < len(triangle[x])```, the minimum path sum of the path that ends up on ```(x, y)``` is giv... | 26 | Given a `triangle` array, return _the minimum path sum from top to bottom_.
For each step, you may move to an adjacent number of the row below. More formally, if you are on index `i` on the current row, you may move to either index `i` or index `i + 1` on the next row.
**Example 1:**
**Input:** triangle = \[\[2\],\[... | null |
Most optimized solution, easy to understand || C++ || JAVA || Python | best-time-to-buy-and-sell-stock | 1 | 1 | # Approach\nTo solve this problem, I employed a straightforward approach that iterates through the array of stock prices. At each step, I kept track of the minimum stock price seen so far (`min_price`) and calculated the potential profit that could be obtained by selling at the current price (`prices[i] - min_price`). ... | 250 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
Return _the maximum profit you can achieve from this transaction_. If you ... | null |
O(n) | best-time-to-buy-and-sell-stock | 0 | 1 | \n# Code\n```\nclass Solution:\n def maxProfit(self, prices: List[int]) -> int:\n m=prices[0]\n d=0\n c=0\n for i in range(1,len(prices)):\n m=min(m,prices[i])\n if prices[i]<prices[i-1]:\n c+=1 \n else:\n diff=prices[i]-m\n ... | 1 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
Return _the maximum profit you can achieve from this transaction_. If you ... | null |
Simplest solution using two pointers | best-time-to-buy-and-sell-stock | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def maxProfit(self, prices: List[int]) -> int:\n l,r=0,1\n maxP=0\n\n while r<len(prices):\n if prices[l]<prices[r]:\n profit=prices[r]-prices[l]\n maxP=max(maxP,profit)\n else:\n l=r\n ... | 1 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
Return _the maximum profit you can achieve from this transaction_. If you ... | null |
Day 56 || O(n) time and O(1) space || Easiest Beginner Friendly Sol | best-time-to-buy-and-sell-stock | 1 | 1 | # Intuition of this Problem:\nThe intuition behind the code is to keep track of the minimum stock value seen so far while traversing the array from left to right. At each step, if the current stock value is greater than the minimum value seen so far, we calculate the profit that can be earned by selling the stock at th... | 93 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
Return _the maximum profit you can achieve from this transaction_. If you ... | null |
[VIDEO] Intuitive Visualization and Proof of O(n) Solution | best-time-to-buy-and-sell-stock | 0 | 1 | https://www.youtube.com/watch?v=ioFPBdChabY\n\nA brute force approach would calculate every possible buy-sell combination and would run in O(n^2), but we can reduce this to O(n) by avoiding unncessary computations. The strategy below iterates once for every sell date, and handles two cases:\n1. If buy price < sell pri... | 6 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
Return _the maximum profit you can achieve from this transaction_. If you ... | null |
Easy || 100% || Kadane's Algorithm (Java, C++, Python, JS, Python3) || Min so far | best-time-to-buy-and-sell-stock | 1 | 1 | Given an array prices where prices[i] is the price of a given stock on the ith day.\nmaximize our profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.\nReturn the maximum profit achieve from this transaction. If you cannot achieve any profit, return 0\n\n**Exam... | 133 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
Return _the maximum profit you can achieve from this transaction_. If you ... | null |
Python 3 -> Very simple solution. Explanation added | best-time-to-buy-and-sell-stock | 0 | 1 | **Suggestions to make it better are always welcomed.**\n\nWe might think about using sliding window technique, but obviously we don\'t need subarray here. We just need one value from the given input list. So, this technique is not useful.\n\n**Solution:**\nWe always need to know what is the maxProfit we can make if we ... | 308 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
Return _the maximum profit you can achieve from this transaction_. If you ... | null |
Buy and sell Stocks in Python | best-time-to-buy-and-sell-stock | 0 | 1 | # Intuition\nJust observe your daily stock buy and sell process\n\n# Approach\nwe start from the beginning of list and if any day we see the price has gone down then we take hold i.e. we buy at that price and if the price if higher we just compute the last max profit and store it\n\n# Complexity\n- Time complexity:O(n)... | 0 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
Return _the maximum profit you can achieve from this transaction_. If you ... | null |
basic o(n) solution .not higly optimised. Only for your reference. :p | best-time-to-buy-and-sell-stock | 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. $$O(n)$$ --... | 0 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
Return _the maximum profit you can achieve from this transaction_. If you ... | null |
Easiest One Linear Solution in Python3 | best-time-to-buy-and-sell-stock | 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. -->\nHere\'s the approach used in this code:\n\n1. The accumulate function is used from the itertools library. It calculates the cumulative sums of the input list prices. F... | 4 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
Return _the maximum profit you can achieve from this transaction_. If you ... | null |
Python || 99.54% Faster || T.C. O(n) S.C. O(1) | best-time-to-buy-and-sell-stock | 0 | 1 | ```\nclass Solution:\n def maxProfit(self, prices: List[int]) -> int:\n n=len(prices)\n m1=prices[0]\n m2=0\n for i in range(1,n):\n if m1>prices[i]:\n m1=prices[i]\n if m2<(prices[i]-m1):\n m2=prices[i]-m1\n return m2 ... | 3 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
Return _the maximum profit you can achieve from this transaction_. If you ... | null |
best-time-to-buy-and-sell-stock | best-time-to-buy-and-sell-stock | 0 | 1 | # Code\n```\nclass Solution:\n def maxProfit(self, nums: List[int]) -> int:\n l = [nums[-1]]\n for i in range(len(nums)-2,-1,-1):\n if nums[i]>l[-1]:\n l.append(nums[i])\n else:\n l.append(l[-1])\n l = l[::-1]\n b = 0\n for i,j in... | 1 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
Return _the maximum profit you can achieve from this transaction_. If you ... | null |
Python3 Easy Solution | best-time-to-buy-and-sell-stock | 0 | 1 | \n# Complexity\n- Time complexity:\n0(N)\n\n- Space complexity:\n0(1)\n\n# Code\n```\nclass Solution:\n def maxProfit(self, arr: List[int]) -> int:\n minSoFar=float(\'inf\')\n maxProfit=0\n for i in range(len(arr)):\n minSoFar=min(minSoFar,arr[i])\n maxProfit=max(maxProfit,... | 1 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
Return _the maximum profit you can achieve from this transaction_. If you ... | null |
Python/JS/Java/Go/C++ O(n) by DP // Greedy [+ Visualization] | best-time-to-buy-and-sell-stock-ii | 1 | 1 | Method_#1\nO(n) sol by DP + state machine\n\nDefine "**Hold**" as the state of **holding stock**. (\u6301\u6709\u80A1\u7968)\nDefine "**NotHold**" as the state of **keep cash**. (\u6301\u6709\u73FE\u91D1)\n\nGeneral rule aka recursive relationship.\n\n```\nDP[Hold] \n= max(keep holding stock, or just buy stock & hold t... | 655 | You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
On each day, you may decide to buy and/or sell the stock. You can only hold **at most one** share of the stock at any time. However, you can buy it then immediately sell it on the **same day**.
Find and return _t... | null |
python + easy_sol + 95% | best-time-to-buy-and-sell-stock-ii | 0 | 1 | \n# Code\n```\nclass Solution:\n def maxProfit(self, prices: List[int]) -> int:\n def back(x):\n while x < len(prices) - 1 and prices[x] < prices[x + 1]:\n x += 1\n return x\n mx = j = i = 0\n while i < len(prices) - 1:\n p = 0\n if pric... | 2 | You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
On each day, you may decide to buy and/or sell the stock. You can only hold **at most one** share of the stock at any time. However, you can buy it then immediately sell it on the **same day**.
Find and return _t... | null |
✅Python3✅ Greedy | best-time-to-buy-and-sell-stock-ii | 0 | 1 | # Please Upvote \uD83D\uDE07\n\n## Python3\n```\nclass Solution:\n def maxProfit(self, a: List[int]) -> int:\n ans=0\n x=a[0]\n for i in range(1,len(a)):\n if(x>a[i]):\n x=a[i]\n else:\n ans+=a[i]-x\n x=a[i]\n return ans\n... | 34 | You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
On each day, you may decide to buy and/or sell the stock. You can only hold **at most one** share of the stock at any time. However, you can buy it then immediately sell it on the **same day**.
Find and return _t... | null |
best time to buy and sell stock 2 - python with description | best-time-to-buy-and-sell-stock-ii | 0 | 1 | # Intuition\nIf we can buy and sell on any day, then all we need to check is day i and day i - 1 to see if buying/selling the next day gives us a profit. If it does, do it. Otherwise don\'t.\n\n# Approach\n1. inizialize profit to 0\n2. iterate through prices starting at 2nd value (so we can i-1)\n3. if the ith price is... | 2 | You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
On each day, you may decide to buy and/or sell the stock. You can only hold **at most one** share of the stock at any time. However, you can buy it then immediately sell it on the **same day**.
Find and return _t... | null |
Fantastic Logic----->Python3 | best-time-to-buy-and-sell-stock-ii | 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. $$O(n)$$ --... | 26 | You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
On each day, you may decide to buy and/or sell the stock. You can only hold **at most one** share of the stock at any time. However, you can buy it then immediately sell it on the **same day**.
Find and return _t... | null |
DETAILED EXPLANATION STOCK-II and STOCK-FEE in Python | best-time-to-buy-and-sell-stock-ii | 0 | 1 | \n# Approach\nIn stock related problems, it is very important to grasp some concepts related to how we can maximize our total profit. Let me state a few obvious points:\n1. We need the minimum possible buying price of that stock\n2. We need the maximum possible sell price of that stock\n\nKeeping these two obvious poin... | 2 | You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
On each day, you may decide to buy and/or sell the stock. You can only hold **at most one** share of the stock at any time. However, you can buy it then immediately sell it on the **same day**.
Find and return _t... | null |
Simple Python Solution with 99%beats | best-time-to-buy-and-sell-stock-ii | 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:99.76%\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:72.54%\n<!-- Add your space complexity here, e.g.... | 1 | You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
On each day, you may decide to buy and/or sell the stock. You can only hold **at most one** share of the stock at any time. However, you can buy it then immediately sell it on the **same day**.
Find and return _t... | null |
Python short 1-liner. Functional programming. | best-time-to-buy-and-sell-stock-ii | 0 | 1 | # Approach\n1. For every day $$d_i$$, if $$prices[d_i] < prices[d_{i + 1}]$$ then buy the stock on $$d_i$$ and sell it on $$d_{i + 1}$$ else do nothing.\n\n2. Continue this for each day and return the `sum` of all the profits.\n\n# Complexity\n- Time complexity: $$O(n)$$\n\n- Space complexity: $$O(1)$$\n\nwhere, `n is ... | 5 | You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
On each day, you may decide to buy and/or sell the stock. You can only hold **at most one** share of the stock at any time. However, you can buy it then immediately sell it on the **same day**.
Find and return _t... | null |
with step by step explanation | best-time-to-buy-and-sell-stock-iii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe basic idea is to iterate over the array of stock prices and update four variables:\n\nbuy1 - the minimum price seen so far for the first transaction\nsell1 - the maximum profit seen so far for the first transaction\nbuy2... | 29 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
Find the maximum profit you can achieve. You may complete **at most two transactions**.
**Note:** You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
**Exampl... | null |
Superb Dynamic Logic | best-time-to-buy-and-sell-stock-iii | 0 | 1 | # Dynamic programming\n```\nclass Solution:\n def maxProfit(self, A: List[int]) -> int:\n N=len(A)\n sell=[0]*N\n for _ in range(2):\n buy=-A[0]\n profit=0\n for i in range(1,N):\n buy=max(buy,sell[i]-A[i])\n profit=max(profit,A[i]+b... | 7 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
Find the maximum profit you can achieve. You may complete **at most two transactions**.
**Note:** You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
**Exampl... | null |
O(n)time/BEATS 99.97% MEMORY/SPEED 0ms MAY 2022 | best-time-to-buy-and-sell-stock-iii | 1 | 1 | ```\n```\n\n(Note: This is part of a series of Leetcode solution explanations. If you like this solution or find it useful, ***please upvote*** this post.)\n***Take care brother, peace, love!***\n\n```\n```\n\n```\n```\n\n```\n```\n\nThe best result for the code below is ***0ms / 38.2MB*** (beats 92.04% / 24.00%).\n* *... | 26 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
Find the maximum profit you can achieve. You may complete **at most two transactions**.
**Note:** You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
**Exampl... | null |
🔥Mastering Buying and Selling Stocks || From Recursion To DP || Complete Guide By Mr. Robot | best-time-to-buy-and-sell-stock-iii | 1 | 1 | # \u2753 Buying and Selling Stocks with K Transactions \u2753\n\n\n## \uD83D\uDCA1 Approach 1: Recursive Approach\n\n### \u2728 Explanation\nThe logic and approach for this solution involve recursively exploring the possibilities of buying and selling stocks with a given number of transactions. The function `Recursive`... | 1 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
Find the maximum profit you can achieve. You may complete **at most two transactions**.
**Note:** You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
**Exampl... | null |
Python || DP || Recursion->Space Optimization | best-time-to-buy-and-sell-stock-iii | 0 | 1 | ```\n#Recursion \n#Time Complexity: O(2^n)\n#Space Complexity: O(n)\nclass Solution1:\n def maxProfit(self, prices: List[int]) -> int:\n def solve(ind,buy,cap):\n if ind==n or c==0:\n return 0\n profit=0\n if buy==0: # buy a stock\n take=-prices[i... | 3 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
Find the maximum profit you can achieve. You may complete **at most two transactions**.
**Note:** You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
**Exampl... | null |
Python O(n) by DP // Reduction [w/ Visualization] | best-time-to-buy-and-sell-stock-iii | 0 | 1 | Python sol by DP and state machine\n\n---\n\n**State machine diagram**\n\n\n\n---\n\n**Implementation** by DP:\n\n```\nclass Solution:\n def maxProfit(self, prices: List[int]) -> int:\n \n\t\t\'\'\'\n\t... | 41 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
Find the maximum profit you can achieve. You may complete **at most two transactions**.
**Note:** You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
**Exampl... | null |
Easy O(n) solution, no DP, Python | best-time-to-buy-and-sell-stock-iii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe first split the question to 3 parts: at most 2 trades == max(0 trade, 1 trade, 2 trades)\n\nmaximum profit for doing exactly 0 trade is easy (it will be 0)\nmaximum profit for doing exactly 1 trade is easy (can be calculate in linear t... | 3 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
Find the maximum profit you can achieve. You may complete **at most two transactions**.
**Note:** You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
**Exampl... | null |
[Python 3] Most Intuitive Approach | best-time-to-buy-and-sell-stock-iii | 0 | 1 | # Code\n```\nclass Solution:\n def maxProfit(self, prices: List[int]) -> int:\n initial = 0\n bought1 = -inf\n sold1 = 0\n bought2 = -inf\n sold2 = 0\n for x in prices:\n sold2 = max(bought2 + x, sold2)\n bought2 = max(sold1 - x, bought2)\n s... | 3 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
Find the maximum profit you can achieve. You may complete **at most two transactions**.
**Note:** You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
**Exampl... | null |
Python3 Easy Solution with Comments | best-time-to-buy-and-sell-stock-iii | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def maxProfit(self, prices: List[int]) -> int:\n buy1 = buy2 = float("inf")\n profit1 = profit2 = 0\n \n # Iterate through the prices and update the maximum profits\n for price in prices:\n # Update the price at which we should buy the ... | 4 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
Find the maximum profit you can achieve. You may complete **at most two transactions**.
**Note:** You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
**Exampl... | null |
MOST OPTIMIZED PYTHON SOLUTION | best-time-to-buy-and-sell-stock-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: $$O(N*2*3)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(N*2*3)$$\n<!-- Add your space complex... | 2 | You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
Find the maximum profit you can achieve. You may complete **at most two transactions**.
**Note:** You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
**Exampl... | null |
A very concise python3 solution | binary-tree-maximum-path-sum | 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 | A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.
The **path sum** of a path is the sum of the node's values in the path.
... | null |
CodeDominar solution | binary-tree-maximum-path-sum | 0 | 1 | \n# Code\n```\nclass Solution:\n def maxPathSum(self, root: Optional[TreeNode]) -> int: \n max_ans = float(\'-inf\')\n def helper(root):\n nonlocal max_ans\n if not root:\n return 0\n l = helper(root.left)\n r = helper(root.right)\n ... | 2 | A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.
The **path sum** of a path is the sum of the node's values in the path.
... | null |
Recursive approach✅ | O( n)✅ | Python and C++(Step by step explanation)✅ | binary-tree-maximum-path-sum | 0 | 1 | # Intuition\nThe problem involves finding the maximum path sum in a binary tree. A path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections.\n\n# Approach\nThe approach is to perform a Depth-First Search (DFS) traversal of the binary tree. At each node,... | 5 | A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.
The **path sum** of a path is the sum of the node's values in the path.
... | null |
Self-Explanatory Python Code, Check Concise Version In The Comment! | binary-tree-maximum-path-sum | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nmaxlen gives return the max length from the node down to leaves\n\nnodeMax is the max length joining two branches from left and right, and has to include the current node.val. Basically, the path is like this ^\n\ndfs is just looping thro... | 0 | A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.
The **path sum** of a path is the sum of the node's values in the path.
... | null |
Recursive O(n) solution | binary-tree-maximum-path-sum | 0 | 1 | # Intuition\r\nThe first thing tha comes to mind is using recursion. This is correct but we must be careful. \r\n\r\n# Approach\r\nEach node has its best path, that will be passed up to the father, thanks to recursion. But what are the best path starting from node?\r\n\r\nThere are three possibilities:\r\n\r\n- **value... | 2 | A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.
The **path sum** of a path is the sum of the node's values in the path.
... | null |
Easy Python Recursion with Explanation | Beats 92% | binary-tree-maximum-path-sum | 0 | 1 | # Intuition\nAt each node either that (node) or (node + left subtree) or (node + right subtree) or (left subtree + node + rightsubtree) is the maximum that we can get.\n\nBut, if we reach that node either we can stay ther, or go to left side or right side, we can not come from some other node and then traverse the com... | 2 | A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.
The **path sum** of a path is the sum of the node's values in the path.
... | null |
Most effective and simple recursive solution with explanation | binary-tree-maximum-path-sum | 1 | 1 | \n\n# Approach\n- **Recursive Function (findMaxPathSum):** For each node, calculate the maximum path sum that includes the current node. This sum is the maximum of either the left subtree\'s sum or the right subtree\'s sum, plus the current node\'s value. Update maxi with the maximum of the current maxi value and the s... | 1 | A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.
The **path sum** of a path is the sum of the node's values in the path.
... | null |
Solution | binary-tree-maximum-path-sum | 1 | 1 | ```C++ []\nclass Solution {\npublic:\n int maxPathSum(TreeNode* root) {\n int res = INT_MIN;\n find_max(root, res);\n return res;\n }\n\n int find_max(TreeNode* root, int& res) {\n if (root == nullptr) return 0;\n\n int left_max = find_max(root->left, res);\n int right... | 4 | A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.
The **path sum** of a path is the sum of the node's values in the path.
... | null |
✅ Explained - Simple and Clear Python3 Code✅ | binary-tree-maximum-path-sum | 0 | 1 | The logic for finding the maximum path sum in a binary tree involves traversing the tree in a post-order manner.\n\nAt each node, we consider three scenarios: \n1. the maximum path sum passing through the current node and including its left child\n2. the maximum path sum passing through the current node and including i... | 7 | A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.
The **path sum** of a path is the sum of the node's values in the path.
... | null |
O(n)time/BEATS 99.97% MEMORY/SPEED 0ms MAY 2022 | binary-tree-maximum-path-sum | 1 | 1 | ```\n```\n\n(Note: This is part of a series of Leetcode solution explanations. If you like this solution or find it useful, ***please upvote*** this post.)\n***Take care brother, peace, love!***\n\n```\n```\n\n```\n```\n\n```\n```\n\nThe best result for the code below is ***0ms / 38.2MB*** (beats 92.04% / 24.00%).\n* *... | 20 | A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.
The **path sum** of a path is the sum of the node's values in the path.
... | null |
DFS - Fast and Easy Solution O(N) | binary-tree-maximum-path-sum | 0 | 1 | \n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Definition for a binary tree node.\n# class TreeNode:\n# def __init__(self, val=0, left=None, right=None):\n# s... | 2 | A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.
The **path sum** of a path is the sum of the node's values in the path.
... | null |
[VIDEO] Step-by-Step Visualization of Two Pointer Solution | valid-palindrome | 0 | 1 | https://www.youtube.com/watch?v=4rzENqwlVU8\n\nTo test if a string is a palindrome, we\'ll use a two pointer approach that starts at the ends of the string. At each step, if the characters at the two pointers are equal, then we\'ll move both pointers inwards and compare the next set of characters. Since a palindrome ... | 8 | A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.
Given a string `s`, return `true` _if it is a **palindrome**, or_ `false` _otherwise_... | null |
Easy Python Solution || Valid Palindrome? Let's Find it out..... | valid-palindrome | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal of this function is to determine whether a given string s is a palindrome or not. A palindrome is a string that reads the same backward as forward, ignoring spaces, punctuation, and capitalization.\n# Approach\n<!-- Describe your... | 4 | A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.
Given a string `s`, return `true` _if it is a **palindrome**, or_ `false` _otherwise_... | null |
Valid Palindrome in Python3 | valid-palindrome | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\n#### Here\'s a breakdown of how the method works:\n\n1. new = ("".join(i for i in s if i.isalnum())).lower(): This line of code takes the input string s and performs the following operations:\n\n- It iterates through each character in the input string... | 2 | A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.
Given a string `s`, return `true` _if it is a **palindrome**, or_ `false` _otherwise_... | null |
[Python 3] Two solutions || beats 99% || 33ms 🥷🏼 | valid-palindrome | 0 | 1 | ```python3 []\nclass Solution:\n def isPalindrome(self, s: str) -> bool:\n s = [c.lower() for c in s if c.isalnum()]\n return all (s[i] == s[~i] for i in range(len(s)//2))\n```\n```python3 []\nclass Solution:\n def isPalindrome(self, s: str) -> bool:\n i, j = 0, len(s) - 1\n while i < ... | 39 | A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.
Given a string `s`, return `true` _if it is a **palindrome**, or_ `false` _otherwise_... | null |
Python || 98% Beats ||| Simple code | valid-palindrome | 0 | 1 | # Your upvote is my motivation!\n\n\n# Code\n```\n<!-- Optimize Code -->\n\nclass Solution:\n def isPalindrome(self, s: str) -> bool:\n s1 = \'\'\n for c in s.lower():\n if c.isalnum():\n s1 += c\n\n return True if s1==s1[::-1] else False\n\n============================... | 20 | A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.
Given a string `s`, return `true` _if it is a **palindrome**, or_ `false` _otherwise_... | null |
2 Simple Python Solutions (Two Pointers approach (Beats 99.99%)) | valid-palindrome | 0 | 1 | # Intuition\nThe code aims to determine whether a given string is a palindrome or not. A palindrome is a string that reads the same forwards and backwards. The code uses two pointers, l and r, to traverse the string from the beginning and end simultaneously. It skips non-alphanumeric characters and compares the corresp... | 39 | A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.
Given a string `s`, return `true` _if it is a **palindrome**, or_ `false` _otherwise_... | null |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.