<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 solve(vector<int>& nums , int target , int i , vector<int>&temp ,set<vector<int>>&ans ){
    if( target == 0){
        ans.insert(temp); 
        return ; 
    }

    if(target < 0 || i >= nums.size()){
        return ; 
    }

    if(nums[i] <= target ){
             temp.push_back(nums[i]) ;
         solve(nums , target - nums[i] , i+1 ,temp , ans ) ; 
               temp.pop_back() ; 

    }
    while(i < nums.size()-1 && nums[i] == nums[i+1]){
          i++ ; 
    }
        solve(nums , target  , i+1 ,temp , ans ) ; 

}
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>>ans  ;  set<vector<int>>st  ; 
        int i = 0 ; 
        vector<int>temp ; 
          sort(candidates.begin() , candidates.end()) ;
        solve(candidates , target , i , temp , st ) ; 

        for(auto it : st ){
            ans.push_back(it) ; 
        }

        return ans  ; 
    }
};
13728