<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>> ans;
map<vector<int>,int> m;
void calculateSum(vector<int> cand,int target,int i,vector<int> temp,int sum){
    if(target==sum){
        // sort(temp.begin(),temp.end());
        if(!m[temp])
        ans.push_back(temp);
        m[temp]++;
        return;
    }
if(i>=cand.size() || sum>target)
return;

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