Combination Sum LeetCode Solution

Here, We see Combination Sum problem Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc., with different approaches.

Combination Sum LeetCode Solution

Combination Sum LeetCode Solution

Problem Statement ->

Given 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.

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: []

Combination Sum Leetcode Solution C++ ->

class Solution { public: vector<vector<int>> result; vector<vector<int>> combinationSum(vector<int>& candidates, int target) { vector<int> curr; int sum = 0; int n = candidates.size(); comSum(curr, 0, sum, candidates, target, n); return result; } void comSum(vector<int> &curr, int curInd, int sum, vector<int> &candidates, int target, int n){ if(sum == target){ result.push_back(curr); return; } else if(sum > target){ return; } for(int i = curInd; i < n; i++){ curr.push_back(candidates[i]); sum += candidates[i]; comSum(curr, i, sum, candidates, target, n); sum -= candidates[i]; curr.pop_back(); } } };
Code language: C++ (cpp)

Combination Sum Leetcode Solution Java ->

class Solution { public List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> list = new ArrayList<>(); Arrays.sort(candidates); backtrack(list, new ArrayList<>(), candidates, target, 0); return list; } private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] candidates, int remain, int start){ if(remain < 0) return; else if(remain == 0) list.add(new ArrayList<>(tempList)); else{ for(int i = start; i < candidates.length; i++){ tempList.add(candidates[i]); backtrack(list, tempList, candidates, remain - candidates[i], i); // not i + 1 because we can reuse same elements tempList.remove(tempList.size() - 1); } } } }
Code language: Java (java)

Combination Sum Leetcode Solution JavaScript ->

var combinationSum = function(candidates, target) { var buffer = []; var result = []; search(0, target); return result; function search(startIdx, target) { if (target === 0) return result.push(buffer.slice()); if (target < 0) return; if (startIdx === candidates.length) return; buffer.push(candidates[startIdx]); search(startIdx, target - candidates[startIdx]); buffer.pop(); search(startIdx + 1, target); } };
Code language: JavaScript (javascript)

Combination Sum Leetcode Solution Python ->

class Solution(object): def combinationSum(self, candidates, target): result = [] candidates = sorted(candidates) def dfs(remain, stack): if remain == 0: result.append(stack) return for item in candidates: if item > remain: break if stack and item < stack[-1]: continue else: dfs(remain - item, stack + [item]) dfs(target, []) return result
Code language: Python (python)