<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:
    // Insertion sort for small subarrays
    void insertionSort(vector<int>& nums, int low, int high) {
        for (int i = low + 1; i <= high; ++i) {
            int key = nums[i];
            int j = i - 1;
            while (j >= low && nums[j] > key) {
                nums[j + 1] = nums[j];
                j--;
            }
            nums[j + 1] = key;
        }
    }

    // Merge function to merge two sorted halves
    void merge(vector<int>& nums, int low, int mid, int high) {
        int n1 = mid - low + 1;
        int n2 = high - mid;
        
        // Create temporary arrays
        vector<int> left(n1);
        vector<int> right(n2);
        
        // Copy data to temporary arrays
        for (int i = 0; i < n1; ++i)
            left[i] = nums[low + i];
        for (int j = 0; j < n2; ++j)
            right[j] = nums[mid + 1 + j];
        
        // Merge the temporary arrays back into nums[low..high]
        int i = 0, j = 0, k = low;
        while (i < n1 && j < n2) {
            if (left[i] <= right[j]) {
                nums[k++] = left[i++];
            } else {
                nums[k++] = right[j++];
            }
        }
        
        // Copy the remaining elements of left[], if any
        while (i < n1) {
            nums[k++] = left[i++];
        }
        
        // Copy the remaining elements of right[], if any
        while (j < n2) {
            nums[k++] = right[j++];
        }
    }

    // Merge Sort function
    void mergeSort(vector<int>& nums, int low, int high) {
        if (low < high) {
            if (high - low <= 10) {  // Use insertion sort for small arrays
                insertionSort(nums, low, high);
            } else {
                int mid = low + (high - low) / 2;
                mergeSort(nums, low, mid);
                mergeSort(nums, mid + 1, high);
                merge(nums, low, mid, high);
            }
        }
    }

    // Main function to sort the array
    vector<int> sortArray(vector<int>& nums) {
        mergeSort(nums, 0, nums.size() - 1);
        return nums;
    }
};

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