<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 mergeTwoVectors(vector<int> &nums, int s, int e, int mid)
{
    int j = mid + 1, k = 0, i = s;
    vector<int> ans(e - s + 1);
    while ((i <= mid) && (j <= e))
    {
        if (nums[i] < nums[j])
        {
            ans[k] = nums[i];
            i++;
        }
        else
        {
            ans[k] = nums[j];
            j++;
        }
        k++;
    }
    while (i <= mid)
    {
        ans[k] = nums[i];
        i++;
        k++;
    }
    while (j <= e)
    {
        ans[k] = nums[j];
        j++;
        k++;
    }
    for (int i = s; i <= e; i++)
    {
        nums[i] = ans[i - s];
    }
}

void mergeSort(vector<int> &nums, int s, int e)
{
    if (s >= e)
    {
        return;
    }

    int mid = (s + e) / 2;
    mergeSort(nums, s, mid);
    mergeSort(nums, mid + 1, e);

    mergeTwoVectors(nums, s, e, mid);
}
public:
    vector<int> sortArray(vector<int>& nums) {
        vector<int> ans;
    for (int i = 0; i < nums.size(); i++)
    {
        ans.push_back(nums[i]);
    }

    mergeSort(ans, 0, ans.size() - 1);

    return ans;
    }
};
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