Last updated on February 22nd, 2024 at 03:52 am

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

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Combination Sum II LeetCode Solution

Problem Statement

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sum to target.

Example 1:
Input: candidates = [10,1,2,7,6,1,5], target = 8
Output: 
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]

Example 2:
Input: candidates = [2,5,2,1,2], target = 5
Output: 
[
[1,2,2],
[5]
]

Combination Sum II Leetcode Solution C++

class Solution {
public:
    vector<vector<int>> result;
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<int> curr;
        int n = candidates.size();
        sort(candidates.begin(), candidates.end());
        comsum(curr, target, 0, candidates, 0, n);
        return result;
    }
        void comsum(vector<int> &curr, int target, int sum, vector<int> &candidates, int curInd, int n){
        if(target == sum){
            result.push_back(curr);
            return;
        }
        else if(sum>target){
            return;
        }
        
        for(int i = curInd; i < n; i++){
            if(i != curInd && candidates[i]==candidates[i-1]) 
                continue;
            sum += candidates[i];
            curr.push_back(candidates[i]);
            comsum(curr, target, sum, candidates, i+1, n);
            sum -= candidates[i];
            curr.pop_back();
        }        
    }
};
Code language: C++ (cpp)

Combination Sum II Leetcode Solution Java

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> smallAns = new ArrayList<>();
        
        Arrays.sort(candidates);
        combinationSum2(candidates, target, 0, smallAns, res);
        return res;        
    }
        public int combinationSum2(int[] arr, int tar, int idx, List<Integer> smallAns, List<List<Integer>> res) {
        if (tar == 0) {
            ArrayList<Integer> base = new ArrayList<>(smallAns);
            res.add(base);
            return 1;
        }
        boolean[] visited = new boolean[50];
        int count = 0;
        for (int i = idx; i < arr.length; ++i) {
            if (!visited[arr[i]] && tar - arr[i] >= 0) {
                
                visited[arr[i]] = true;
                smallAns.add(arr[i]);
                count += combinationSum2(arr, tar - arr[i], i + 1, smallAns, res);
                smallAns.remove(smallAns.size() - 1);
            }
        }
        return count;
    }
}
Code language: Java (java)

Combination Sum II Leetcode Solution JavaScript

var combinationSum2 = function(candidates, target) {
    if (!candidates) {
        return [];
    }
    if (target === 0) {
        return [[]];
    }
    candidates.sort((a,b) => { return a - b});
    let paths = [];
    let find = function (t, p, i) {
        if (t === 0) {
            paths.push(p);
            return;
        } else {
            while (i < candidates.length && t - candidates[i] >= 0) {
                find(t - candidates[i], [...p, candidates[i]], i + 1)
                i++;
                while (candidates[i - 1] === candidates[i]) {
                    i++;
                }
            }
        }     
    }
    find (target, [], 0);
    return paths;    
};
Code language: JavaScript (javascript)

Combination Sum II Leetcode Solution Python

class Solution(object):
    def combinationSum2(self, candidates, target):
        candidates.sort()                      
        result = []
        self.combine_sum_2(candidates, 0, [], result, target)
        return result
    def combine_sum_2(self, nums, start, path, result, target):
        if not target:
            result.append(path)
            return
        for i in xrange(start, len(nums)):
            if i > start and nums[i] == nums[i - 1]:
                continue
            if nums[i] > target:
                break
            self.combine_sum_2(nums, i + 1, path + [nums[i]], 
                result, target - nums[i])
Code language: Python (python)