<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:
    void merge(vector<int> &nums,
               int left_first,
               int left_second,
               int right_first,
               int right_second
    ) {
        int size = right_second - left_first;
        vector<int> v;
        int i = left_first, j = right_first;
        int n1 = left_second;
        int n2 = right_second;
        int x, y;
        while (i < n1 && j < n2) {
            x = nums[i];
            y = nums[j];
            if (x < y) {
                v.push_back(x);
                ++i;
            } else {
                v.push_back(y);
                ++j;
            }
        }
        if (i == n1 && j != n2) {
            for (; j < n2; ++j) {
                v.push_back(nums[j]);
            }
        } else if (j == n2 && i != n1) {
            for (; i < n1; ++i) {
                v.push_back(nums[i]);
            }
        }
        for (i = 0; i < size; ++i) {
            nums[left_first + i] = v[i];
        }

    }

    vector<int> sortArray(vector<int> &nums) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);


        int n = nums.size();
        int
                left_first,
                left_second,
                right_first,
                right_second;
        for (int size = 1; size < n; size *= 2) {
            for (int i = 0; i < n; i += 2 * size) {
                left_first = i;
                left_second = i + size;
                if (left_second < n ) {
                    right_first = i + size;
                    right_second = min(i + 2 * size, n);
                    merge(
                            nums,
                            left_first, left_second,
                            right_first, right_second
                    );
                }
            }
        }


        return nums;
    }
};
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