In this post, you will know how to solve the Combination Sum Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.Combination Sum Leetcode SolutionOne more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.ProblemGiven an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.The test cases are generated such that the number of unique combinations that sum up to target is less than 150 combinations for the given input.Example 1:Input: candidates = [2,3,6,7], target = 7 Output: [[2,2,3],[7]] Explanation: 2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times. 7 is a candidate, and 7 = 7. These are the only two combinations. Example 2:Input: candidates = [2,3,5], target = 8 Output: [[2,2,2,2],[2,3,3],[3,5]]Example 3:Input: candidates = [2], target = 1 Output: []Constraints:1 <= candidates.length <= 302 <= candidates[i] <= 40All elements of candidates are distinct.1 <= target <= 500Combination Sum Leetcode Solutions in Pythonclass Solution: def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]: ans = [] def dfs(s: int, target: int, path: List[int]) -> None: if target < 0: return if target == 0: ans.append(path.clone()) return for i in range(s, len(candidates)): path.append(candidates[i]) dfs(i, target - candidates[i], path) path.pop() candidates.sort() dfs(0, target, []) return ans Combination Sum Leetcode Solutions in CPPclass Solution { public: vector<vector<int>> combinationSum(vector<int>& candidates, int target) { vector<vector<int>> ans; sort(begin(candidates), end(candidates)); dfs(candidates, 0, target, {}, ans); return ans; } private: void dfs(const vector<int>& A, int s, int target, vector<int>&& path, vector<vector<int>>& ans) { if (target < 0) return; if (target == 0) { ans.push_back(path); return; } for (int i = s; i < A.size(); ++i) { path.push_back(A[i]); dfs(A, i, target - A[i], move(path), ans); path.pop_back(); } } }; Combination Sum Leetcode Solutions in Javaclass Solution { public List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> ans = new ArrayList<>(); Arrays.sort(candidates); dfs(0, candidates, target, new ArrayList<>(), ans); return ans; } private void dfs(int s, int[] candidates, int target, List<Integer> path, List<List<Integer>> ans) { if (target < 0) return; if (target == 0) { ans.add(new ArrayList<>(path)); return; } for (int i = s; i < candidates.length; ++i) { path.add(candidates[i]); dfs(i, candidates, target - candidates[i], path, ans); path.remove(path.size() - 1); } } } Note: This problem Combination Sum is generated by Leetcode but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.Next: Jump Game Leetcode Solution Post navigationCount and Say Leetcode Solution Jump Game Leetcode Solution