<p>Given an array of integers <code>nums</code>, sort the array in ascending order and return it.</p>

<p>You must solve the problem <strong>without using any built-in</strong> functions in <code>O(nlog(n))</code> time complexity and with the smallest space complexity possible.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,2,3,1]
<strong>Output:</strong> [1,2,3,5]
<strong>Explanation:</strong> After sorting the array, the positions of some numbers are not changed (for example, 2 and 3), while the positions of other numbers are changed (for example, 1 and 5).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,1,1,2,0,0]
<strong>Output:</strong> [0,0,1,1,2,5]
<strong>Explanation:</strong> Note that the values of nums are not necessairly unique.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-5 * 10<sup>4</sup> &lt;= nums[i] &lt;= 5 * 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    vector<int> sortArray(vector<int>& nums) {
        quick_sort(nums);
        return nums;
    }

    void quick_sort(vector<int>& nums){
        stack<pair<int, int>> stk;
        stk.push({0, nums.size()-1});

        while(!stk.empty()){
            auto [low, high] = stk.top();
            stk.pop();

            if(low >= high){
                continue;
            }

            int pivot = nums[low];
            int lt = low;
            int gt = high;
            int i = low + 1;

            while(i <= gt){
                if(nums[i] < pivot){
                    swap(nums[lt++], nums[i++]);
                }

                else if(nums[i] > pivot){
                    swap(nums[gt--], nums[i]);
                }

                else{
                    i++;
                }
            }

            stk.push({low, lt-1});
            stk.push({gt+1, high});
        }
    }
};
74041