<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 backtrack(vector<vector<int>>& sol, vector<int> cursol, int target, int cursum, int ind, vector<int>& counts) {
        if (cursum == target) {
            sol.push_back(cursol);
            return;
        }
        if (ind == target+1) return;
        if (cursum > target) return;
        
        for (int i = 0; i <= counts[ind]; i++) {
            if (target < cursum) break;
            backtrack(sol, cursol, target, cursum, ind+1, counts);
            cursol.push_back(ind);
            cursum += ind;
        }
    }

    vector<vector<int>> combinationSum2(vector<int>& A, int target) {
        int n = A.size();
        vector<int> counts(target+1);
        for (auto it : A) {
            if (it <= target) counts[it]++;
        }
        vector<vector<int>> sol;
        vector<int> cursol = {};
        backtrack(sol, cursol, target, 0, 0, counts);
        return sol;
    }
};
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