Combination Sum II [Medium] — Backtracking with Duplicate Skip

Problem Statement

Like Combination Sum but each number can only be used once, and the candidates may contain duplicates. Return unique combinations.

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

Intuition

Sort candidates. In backtracking, skip candidates[i] if i > start and candidates[i] == candidates[i-1] to avoid duplicate combinations at the same recursion level.

Solutions

C++

void dfs(vector<int>& c, int rem, int start, vector<int>& cur, vector<vector<int>>& res) {
    if (rem == 0) { res.push_back(cur); return; }
    for (int i = start; i < c.size() && c[i] <= rem; i++) {
        if (i > start && c[i] == c[i-1]) continue;
        cur.push_back(c[i]); dfs(c, rem-c[i], i+1, cur, res); cur.pop_back();
    }
}
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
    sort(candidates.begin(), candidates.end());
    vector<vector<int>> res; vector<int> cur;
    dfs(candidates, target, 0, cur, res);
    return res;
}

Python

def combinationSum2(candidates: list[int], target: int) -> list[list[int]]:
    candidates.sort()
    res = []
    def dfs(start, rem, path):
        if rem == 0: res.append(path[:]); return
        for i in range(start, len(candidates)):
            if candidates[i] > rem: break
            if i > start and candidates[i] == candidates[i-1]: continue
            path.append(candidates[i])
            dfs(i+1, rem-candidates[i], path)
            path.pop()
    dfs(0, target, [])
    return res

JavaScript

var combinationSum2 = function(candidates, target) {
    candidates.sort((a,b)=>a-b);
    const res=[];
    function dfs(start,rem,cur){
        if(rem===0){res.push([...cur]);return;}
        for(let i=start;i<candidates.length&&candidates[i]<=rem;i++){
            if(i>start&&candidates[i]===candidates[i-1])continue;
            cur.push(candidates[i]);dfs(i+1,rem-candidates[i],cur);cur.pop();
        }
    }
    dfs(0,target,[]);
    return res;
};

Complexity

Time: O(2^n)
Space: O(n)