<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:

    void recursion (int i,int n ,vector<int>&candidates, int target,vector<int>&temp,vector<vector<int>>&res)
    {
        if(target==0)
        {
            res.push_back(temp);
            return;
        }

        if(target<0)
        return;

        if(i==n)
        return;

        unordered_set<int>s;
        for(int k=i;k<n;k++)
        {
            if(s.find(candidates[k])==s.end())
            {
                s.insert(candidates[k]);
                temp.push_back(candidates[k]);
                recursion(k+1,n,candidates,target-candidates[k],temp,res);
                temp.pop_back();
            }
        }
    }

    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>>res;
        sort(candidates.begin(),candidates.end());

        vector<int>temp;
        int n=candidates.size();
        recursion(0,n,candidates,target,temp,res);
        return res;
    }
};
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