<p>Given a collection of candidate numbers (<code>candidates</code>) and a target number (<code>target</code>), find all unique combinations in <code>candidates</code>&nbsp;where the candidate numbers sum to <code>target</code>.</p>

<p>Each number in <code>candidates</code>&nbsp;may only be used <strong>once</strong> in the combination.</p>

<p><strong>Note:</strong>&nbsp;The solution set must not contain duplicate combinations.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> candidates = [10,1,2,7,6,1,5], target = 8
<strong>Output:</strong> 
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> candidates = [2,5,2,1,2], target = 5
<strong>Output:</strong> 
[
[1,2,2],
[5]
]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;=&nbsp;candidates.length &lt;= 100</code></li>
	<li><code>1 &lt;=&nbsp;candidates[i] &lt;= 50</code></li>
	<li><code>1 &lt;= target &lt;= 30</code></li>
</ul>

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());  // Step 1: Sort the candidates array
        return dfs(candidates, target, 0, 0, {});
    }

private:
    vector<vector<int>> dfs(vector<int>& candidates, int target, int sum, int idx, vector<int> ans) {
        if (sum > target) return {};
        if (sum == target) return {ans};

        vector<vector<int>> sol;

        for (int i = idx; i < candidates.size(); ++i) {
            if (i > idx && candidates[i] == candidates[i - 1]) continue;  // Step 2: Skip duplicates

            ans.push_back(candidates[i]);
            vector<vector<int>> next = dfs(candidates, target, sum + candidates[i], i + 1, ans);
            sol.insert(sol.end(), next.begin(), next.end());
            ans.pop_back();  // Backtrack
        }

        return sol;
    }
};

20441