<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 findSubsets(vector<int>v,int s, int idx, int t,vector<int>&a, vector<vector<int>>&ans)
    {
        if(s==t)
        {
            if(find(ans.begin(),ans.end(),a)==ans.end())
                ans.push_back(a);
            return;
        }
        int n=v.size();
        for(int i=idx+1;i<n;i++)
        {
            if(i>=idx+2 && v[i]==v[i-1])
                continue;
            if(s+v[i]>t)
                break;
            a.push_back(v[i]);
            findSubsets(v,s+v[i],i,t,a,ans);
            a.pop_back();
        }
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>>ans;
        sort(candidates.begin(),candidates.end());
        int n=candidates.size();
        for(int i=0;i<n;i++)
        {
            vector<int>a;
            a.push_back(candidates[i]);
            findSubsets(candidates,candidates[i],i,target,a,ans); 
        }
        return ans;
    }
};
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