<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>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> valid_subsets;

        std::sort(candidates.begin(), candidates.end());

        std::vector<bool> subset_bit_vector(candidates.size(), false);
        _combinationSum2(0, false, subset_bit_vector, 0, candidates, target, valid_subsets);

        return valid_subsets;
    }

    void _combinationSum2(
        int i,
        bool included_previous,
        std::vector<bool>& subset_bit_vector,
        int subset_sum,
        vector<int>& candidates, int target,
        vector<vector<int>>& valid_subsets
        )
    {
        /*
            This check is done before checking if (i >= candidates.size())
            Why ? because subset_sum can be modified just before the jump
            here
        */
        if (subset_sum == target)
        {
            std::vector<int> valid_subset;
            for (int j=0; j < candidates.size(); ++j)
            {
                if (subset_bit_vector[j])
                    valid_subset.push_back(candidates[j]);
            }

            if (!valid_subset.empty())
                valid_subsets.push_back(valid_subset);

            return;
        }

        // check if we've exhausted candidates
        if (i == candidates.size())
            return;
        
        if (subset_sum > target)
            return;

        // include current candidate
        /*
        We can prune the decision tree here for equal elements that are
        adjacent. Suppose a[i] == a[i+1]. When we consider a[i] we have
        a decision tree that continues with whether or not we consider a[i+1]
        When we don't consider a[i], we need not not consider a[i+1]. Why? Because
        this is a repeat of the descision tree that considered a[i] and did not
        consider a[i+1]. So we can prune either of the trees
        */
        if (! ((i > 0) && (candidates[i] == candidates[i-1]) && (!included_previous)) )
        {
            subset_bit_vector[i] = true;
            _combinationSum2(i+1, true, subset_bit_vector, subset_sum + candidates[i],
                candidates, target, valid_subsets
            );
        }

        // exclude current candidate
        subset_bit_vector[i] = false;
        _combinationSum2(i+1, false, subset_bit_vector, subset_sum,
            candidates, target, valid_subsets
        );
    }
};
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