<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 final_ans(vector<vector<int>> &ans, vector<int> &temp, int target, int i, int n, vector<int> c){
        if(target==0){
            ans.push_back(temp);
            return;
        }
        for(int r=i;r<n;r++){
            if (r>i && c[r]==c[r-1])
                continue;
            if (c[i]>target)
                break;
            temp.push_back(c[r]);
            final_ans(ans,temp,target-c[r],r+1,n,c);
            temp.pop_back();
        }
    }
    vector<vector<int>> combinationSum2(vector<int>& c, int t) {
        sort(c.begin(),c.end());
        vector<vector<int>> ans;
        vector<int> temp;
        int n=c.size();
        final_ans(ans,temp,t,0,n,c);
        return ans;
    }
};
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