<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>> getAns(int i,vector<int> temp,int target,vector<int> &candidates,vector<vector<int>> &finalAns){
        int n = candidates.size();
        if(target == 0){
            finalAns.push_back(temp);
            return finalAns;
        } 
        
        if(i == n) return finalAns;

            for(int j = i; j<n; j++){
                if(j>i && candidates[j] == candidates[j-1]) continue;
                if(candidates[j]>target) break;
                temp.push_back(candidates[j]);
                getAns(j+1,temp,target-candidates[j],candidates,finalAns);
                temp.pop_back();
            } 
        return finalAns;
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<vector<int>> finalAns;
        vector<int> temp;
        return getAns(0,temp,target,candidates,finalAns);

    }
};
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