<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:
    // time/space: O(nlogn)/O(n)
    vector<int> sortArray(vector<int>& nums) {
        int n = nums.size();

        for (int len = 1; len < n; len *= 2) {
            for (int i = 0, j = len; (i < n) && (j < n); i += (2 * len), j += (2 * len)) {
                merge(nums, i, i + len - 1, j, min(n - 1, j + len - 1));
            }
        }

        return nums;
    }
private:
    void merge(vector<int>& nums, int i, int boundi, int j, int boundj) {
        int l = i, r = boundj;
        vector<int> temp(r - l + 1);
        int k = 0;
        while ((i <= boundi) && (j <= boundj)) {
            if (nums[i] <= nums[j]) {
                temp[k] = nums[i];
                k++, i++;
            }
            else {
                temp[k] = nums[j];
                k++, j++;
            }
        }
        while (i <= boundi) temp[k++] = nums[i++];
        while (j <= boundj) temp[k++] = nums[j++];
        for (int x = 0; x < temp.size(); x++) nums[l + x] = temp[x];
    }
};
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