Leetcode Pascal’s Triangle II problem solution YASH PAL, 31 July 2024 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. Topics we are covering Toggle Problem solution in Python.Problem solution in Java.Problem solution in C++.Problem solution in C. 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 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