Leetcode Pascal’s Triangle II problem solution YASH PAL, 31 July 202419 January 2026 In this Leetcode Pascal’s Triangle II problem solution, we have given an integer rowIndex, return the rowIndexth (0-indexed) row of Pascal’s triangle. In Pascal’s triangle, each number is the sum of the two numbers directly above it.Leetcode Pascal’s Triangle II problem solution in Python.class Solution: def getRow(self, rowIndex: int) -> List[int]: if rowIndex == 0: return [1] if rowIndex == 1: return [1, 1] if rowIndex == 2: return [1, 2, 1] ans = [1] * (rowIndex + 1) prev = [1, 2, 1] for i in range(3, rowIndex + 1): for j in range(1, i): ans[j] = prev[j] + prev[j - 1] prev = ans[:] return ans Pascal’s Triangle II problem solution in Java.public List<Integer> getRow(int rowIndex) { List<Integer> result = new ArrayList<>(); result.add(1); if(rowIndex==0) return result; List<Integer> prev = getRow(rowIndex-1); for(int i=0; i<prev.size()-1;i++){ result.add(prev.get(i)+prev.get(i+1)); } result.add(1); return result; } Problem solution in C++.class Solution { public: vector<int> getRow(int rowIndex) { ++rowIndex; std::vector<int> res(rowIndex, 1); for(int i{1}; i < rowIndex; ++i) res[i] = (int64_t) res[i-1] * (rowIndex-i) / i; return res; } }; Problem solution in C.int* getRow(int rowIndex, int* returnSize) { int *result = malloc((rowIndex + 1) * sizeof(int)); int i, j, new, old = 1; *returnSize = rowIndex + 1; for (i = 0; i <= rowIndex; i++) { result[i] = 1; for (j = 1; j < i; j++) { new = result[j]; result[j] += old; old = new; } } return result; } coding problems solutions Leetcode Problems Solutions Leetcode