<p>Given an array <code>nums</code> of distinct integers, return <em>all the possible permutations</em>. You can return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [0,1]
<strong>Output:</strong> [[0,1],[1,0]]
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> nums = [1]
<strong>Output:</strong> [[1]]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 6</code></li>
	<li><code>-10 &lt;= nums[i] &lt;= 10</code></li>
	<li>All the integers of <code>nums</code> are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        vector<pair<int, bool>> used_nums;
        int len = nums.size();
        for (int i = 0; i < len; i++) {
            used_nums.push_back({nums[i], true});
        }

        vector<vector<int>> ans;
        vector<int> curr;
        permutate(ans, curr, used_nums, len);

        return ans;
    }

    void permutate(vector<vector<int>>& ans, vector<int> curr, vector<pair<int, bool>> used_nums, int& len)
    {
        if (curr.size() == len) {
            ans.push_back(curr);
            return;
        }

        for (int i = 0; i < len; i++) {
            if (used_nums[i].second == true) {
                vector<pair<int, bool>> new_used_nums = used_nums;
                new_used_nums[i].second = false;

                vector<int> new_curr = curr;
                new_curr.push_back(used_nums[i].first);

                permutate(ans, new_curr, new_used_nums, len);
            }
        }
    }
};
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