<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>>ans;
vector<int>a;
void solve(int i , int t , int sum , vector<int>&c){
    if(sum==t){
        ans.push_back(a);
        return ;
    }
    if(i>= c.size() || sum>t) return;
    int in = upper_bound(c.begin() , c.end() ,c[i]) - c.begin();
    solve(in , t , sum , c);
    a.push_back(c[i]);
    solve(i+1 ,t ,sum+c[i] ,c);
    a.pop_back();

}
    vector<vector<int>> combinationSum2(vector<int>& c, int t){

        sort(c.begin() , c.end());
        solve(0 , t ,0 ,c);
        return ans;
        
    }
};
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