<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 helper(vector<int>& a, vector<vector<int>>&ans, vector<int>temp, int tgt, int idx){
    
        if(tgt==0){  // condition hit
            ans.push_back(temp);
        return;
        }
    for(int i= idx; i<a.size() ;i++){
        if(i!=idx && a[i]==a[i-1]) continue; // save us from same occurence of an element and from the same index condition
        if(a[idx]<=tgt){ // for the numbers which can be picked and which aren't
        temp.push_back(a[i]); //pushing into temp ans
        helper(a, ans, temp, tgt-a[i], i+1); // i+1 because we do not want same element addition each time
        temp.pop_back(); //removing the same added element in same call when it"s succedded calls are satisfied
    }
    }
}
    vector<vector<int>> combinationSum2(vector<int>& a, int tgt) {
        sort(a.begin(), a.end());
        vector<int> temp;
        vector<vector<int>> ans;
        
        helper(a, ans, temp, tgt, 0);
        return ans;
    }
};
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