<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;
    int n ,T; 
    void solve(vector<int>&v,int sum ,int idx,vector<int>&cur)
    {
        if(sum>=T)
        {
            if(sum==T) result.push_back(cur);
            return;
        }
            
        if(idx==n)
        {
            return;
        }
        for(int i = idx ; i< n ; i++ )
        {
            if(i>idx && v[i]==v[i-1]) continue;
            // if(i==0 || v[i]!=v[i-1])
            // {
                cur.push_back(v[i]);
                solve(v,sum+v[i],i+1,cur);
                cur.pop_back();
            // }
        }
    }
    

    vector<vector<int>> combinationSum2(vector<int>& v, int target) 
    {
        n = v.size()    ;
        map<int,int>mp;
        for(auto u : v) mp[u]++;
        T= target;
        vector<int>cur;
        sort(v.begin(),v.end());
        solve(v,0,0,cur);
        return result;
    }
};
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