<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 {
    void merge(int* arr, int* arr1, int* arr2, int count1, int count2) {
        int idx=0, jdx=0, order = 0;
        while(idx < count1 && jdx < count2) {
            arr[order++] = arr1[idx] < arr2[jdx] ? arr1[idx++] : arr2[jdx++];
        }
        if (idx == count1) { while (jdx < count2) arr[order++] = arr2[jdx++]; }
        else { while (idx < count1) arr[order++] = arr1[idx++]; }
    };
    void mergesort(int* arr, int count) {

        if (count == 1) {
            return;
        } else if (count == 2) {
            if (arr[0] > arr[1]) swap(arr[0], arr[1]);
            return;
        }
        int count1 = count/2;
        int count2 = count1 + count % 2;

        int* arr1 = new int[count1];
        int* arr2 = new int[count2];

        memcpy(arr1, arr, count1*sizeof(int));
        memcpy(arr2, arr+count1, count2*sizeof(int));
        mergesort(arr1, count1);
        mergesort(arr2, count2);
        merge(arr, arr1, arr2, count1, count2);
        delete[] arr2;
        delete[] arr1;
    };


public:
    vector<int> sortArray(vector<int>& nums) {
        mergesort(&nums[0], nums.size());
        return nums;
    }
};
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