<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>> result;
    void solve(vector<int>& candidates, int target,int idx,vector<int> curr){
        if(target<0) return;
        if(target==0){
            result.push_back(curr);
        }
        for(int i=idx;i<candidates.size();i++){
            if(i>idx && candidates[i]==candidates[i-1]){
                continue;
            }
            curr.push_back(candidates[i]);  //DO
            solve(candidates,target-candidates[i],i+1,curr); //EXPLORE
            curr.pop_back();    //UNDO
        }
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<int> curr;
        sort(begin(candidates),end(candidates));
        solve(candidates,target,0,curr);
        return result;
    }
};
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