<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>> func(vector<int> cand, int ind, int tar, vector<vector<int>> &ans, vector<int> &comb){
        // Base case: If the target becomes zero, we've found a valid combination
        if (tar == 0) {
            ans.push_back(comb);  // Add the current combination to the answer list
            return ans;  // Return the answer list (though not necessary since ans is passed by reference)
        }

        // Iterate over the candidates starting from the current index
        for (int i = ind; i < cand.size(); i++) {
            if (i > ind && cand[i] == cand[i - 1]) continue;// Skip duplicates: If the current candidate is the same as the previous one at this level, skip it
            if (tar < cand[i]) break;// If the current candidate exceeds the remaining target, no need to continue
            comb.push_back(cand[i]);// Include the current candidate in the combination
            func(cand, i + 1, tar - cand[i], ans, comb);// Recursively call func with the next index and the updated target
            comb.pop_back();// Backtrack: Remove the last added candidate to explore other possibilities
        }
        return ans;  // Return the answer list (though it's not necessary)
    }

    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());// Sort the candidates to handle duplicates easily
        vector<vector<int>> ans;  // This will store all the valid combinations
        vector<int> comb;  // This will store the current combination being explored
        func(candidates, 0, target, ans, comb); // Start the recursive function to explore all combinations
        return ans;  // Return the list of all valid combinations
    }
};
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