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In this post, you will find the solution for the **Pascal’s Triangle** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Pascal’s Triangle– LeetCode Problem

Pascal's Triangle– LeetCode Problem

**Problem:**

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 = 5Output:[[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]]

**Example 2:**

Input:numRows = 1Output:[[1]]

**Constraints:**

`1 <= numRows <= 30`

Pascal's Triangle– LeetCode Solutions

Pascal's Triangle Solution in C++:

class Solution { public: vector<vector<int>> generate(int numRows) { vector<vector<int>> ans; for (int i = 0; i < numRows; ++i) ans.push_back(vector<int>(i + 1, 1)); for (int i = 2; i < numRows; ++i) for (int j = 1; j < ans[i].size() - 1; ++j) ans[i][j] = ans[i - 1][j - 1] + ans[i - 1][j]; return ans; } };

Pascal's Triangle Solution in Java:

class Solution { public List<List<Integer>> generate(int numRows) { List<List<Integer>> ans = new ArrayList<>(); for (int i = 0; i < numRows; ++i) { Integer[] temp = new Integer[i + 1]; Arrays.fill(temp, 1); ans.add(Arrays.asList(temp)); } for (int i = 2; i < numRows; ++i) for (int j = 1; j < ans.get(i).size() - 1; ++j) ans.get(i).set(j, ans.get(i - 1).get(j - 1) + ans.get(i - 1).get(j)); return ans; } }

Pascal's Triangle Solution in Python:

class Solution: def generate(self, numRows: int) -> List[List[int]]: ans = [] for i in range(numRows): ans.append([1] * (i + 1)) for i in range(2, numRows): for j in range(1, len(ans[i]) - 1): ans[i][j] = ans[i - 1][j - 1] + ans[i - 1][j] return ans