<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) {
        merge_sort(nums);
        return nums;
    }

    void merge_sort(vector<int> &nums) {
        if (nums.size() <= 1) {
            return;
        }

        int mid = nums.size() / 2;
        vector<int> nums1, nums2;
        for (int i = 0; i < mid; i++) {
            nums1.emplace_back(nums[i]);
        }
        for (int i = mid; i < nums.size(); i++) {
            nums2.emplace_back(nums[i]);
        }

        merge_sort(nums1);
        merge_sort(nums2);

        int idx1 = 0, idx2 = 0;
        while (idx1 < nums1.size() && idx2 < nums2.size()) {
            if (nums1[idx1] < nums2[idx2]) {
                nums[idx1 + idx2] = nums1[idx1];
                idx1++;
            } else {
                nums[idx1 + idx2] = nums2[idx2];
                idx2++;
            }
        }
        while (idx1 < nums1.size()) {
            nums[idx1 + idx2] = nums1[idx1];
            idx1++;
        }
        while (idx2 < nums2.size()) {
            nums[idx1 + idx2] = nums2[idx2];
            idx2++;
        }

        return;
    }
};
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