<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 {
    vector<vector<int>> ans;
    int n;
    vector<int> c;
    void solve(int i, int tar, std::vector<int> temp){
        if(tar==0){
            ans.push_back(temp);
        }
        if(i>=n || tar<0) return;

        int prev=-1;
        for(int j =i; j<n; j++){
            if(prev==c[j]) continue; // this is why we have to sort it so that we can eliminate duplicate elements
            temp.push_back(c[j]);
            solve(j+1,tar-c[j],temp);
            temp.pop_back();
            prev=c[j];
        }
    }
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        n = candidates.size();
        std::sort(candidates.begin(), candidates.end());
        c=candidates;

        std::vector<int> temp;
        solve(0,target,temp);
        return ans;
    }
};
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