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],[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 <= 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<>();
                        cl.add(cands[j]); cl.addAll(l);
                        newList.add(cl);
                    }
                }
            }
            dp.add(newList);
        }
        return dp.get(t-1);
    }
}

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