<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>

#include <algorithm>
#include <climits>
#include <functional>
#include <iostream>
#include <queue>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
static const bool Booster = []() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  return true;
}();

using namespace std;
class Solution {
private:
  void dfs(vector<int> const &candidates, int const &target,
           vector<vector<int>> &combinations, vector<int> &current, int idx) {
    if (target == 0) {
      // Since there are no '0's, just prune the tree
      combinations.push_back(current);
      return;
    }

    if (idx >= candidates.size()) {
      // Out of bounds. Pruned
      return;
    }

    if (target < 0) {
      // Can not ever sum to target. Pruned.
      return;
    }

    // Otherwise, try adding this element
    current.push_back(candidates[idx]);
    dfs(candidates, target - candidates[idx], combinations, current, idx + 1);
    current.pop_back();

    // To skip this element, we need to skip all other elements that are the
    // same as this element. Since the array is sorted, we can skip adjacent
    // elements if they are equal
    while (idx + 1 < candidates.size() &&
           candidates[idx + 1] == candidates[idx]) {
      ++idx;
    }
    dfs(candidates, target, combinations, current, idx + 1);
  }

public:
  vector<vector<int>> combinationSum2(vector<int> &candidates, int target) {
    // To ensure that there are no duplicate combinations, we need to be able to
    // skip numbers that are already in the current combination
    sort(candidates.begin(), candidates.end());

    vector<vector<int>> combinations;
    vector<int> current;
    dfs(candidates, target, combinations, current, 0);

    return combinations;
  }
};

15856