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# Combination Sum LeetCode Solution

## Problem – Combination Sum LeetCode Solution

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.

It is guaranteed 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],]
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 = , target = 1
Output: []
``````

Constraints:

• `1 <= candidates.length <= 30`
• `1 <= candidates[i] <= 200`
• All elements of `candidates` are distinct.
• `1 <= target <= 500`

### Combination Sum LeetCode Solution in C++

``````class Solution {
void combination(vector<int>& candidates, int target, vector<int> currComb, int currSum, int currIndex, vector<vector<int>>& ans){
if(currSum>target) return; //backtrack
if(currSum==target){
ans.push_back(currComb); //store the solution and backtrack
return;
}

for(int i=currIndex; i<candidates.size(); i++){ //try all possible options for the next level
currComb.push_back(candidates[i]); //put 1 option into the combination
currSum+=candidates[i];
combination(candidates, target, currComb, currSum, i, ans); //try with this combination, whether it gives a solution or not.
currComb.pop_back(); //when this option backtrack to here, remove this and go on to the next option.
currSum-=candidates[i];
}

}
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int>> ans;
vector<int> currComb;
combination(candidates, target, currComb, 0, 0, ans);
return ans;
}
};

``````

### Combination Sum LeetCode Solution in Python

``````class Solution(object):
def combinationSum(self, candidates, target):
ret = []
self.dfs(candidates, target, [], ret)
return ret

def dfs(self, nums, target, path, ret):
if target < 0:
return
if target == 0:
ret.append(path)
return
for i in range(len(nums)):
self.dfs(nums[i:], target-nums[i], path+[nums[i]], ret)
``````

### Combination Sum LeetCode Solution in Java

``````public class Solution {
public List<List<Integer>> combinationSum(int[] cands, int t) {
Arrays.sort(cands); // sort candidates to try them in asc order
List<List<List<Integer>>> dp = new ArrayList<>();
for (int i = 1; i <= t; i++) { // run through all targets from 1 to t
List<List<Integer>> newList = new ArrayList(); // combs for curr i
// run through all candidates <= i
for (int j = 0; j < cands.length && cands[j] <= i; j++) {
// special case when curr target is equal to curr candidate
if (i == cands[j]) newList.add(Arrays.asList(cands[j]));
// if current candidate is less than the target use prev results
else for (List<Integer> l : dp.get(i-cands[j]-1)) {
if (cands[j] <= l.get(0)) {
List cl = new ArrayList<>();
}
}
}
}
return dp.get(t-1);
}
}
``````
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