<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:
    void merge(std::vector<int>& nums, int left, int mid, int right) {
    int n1 = mid - left + 1;
    int n2 = right - mid;
    
    // Create temporary arrays
    std::vector<int> L(n1);
    std::vector<int> R(n2);
    
    // Copy data to temporary arrays L[] and R[]
    for (int i = 0; i < n1; ++i)
        L[i] = nums[left + i];
    for (int i = 0; i < n2; ++i)
        R[i] = nums[mid + 1 + i];
    
    // Merge the temporary arrays back into nums[left..right]
    int i = 0, j = 0;
    int k = left;
    while (i < n1 && j < n2) {
        if (L[i] <= R[j]) {
            nums[k] = L[i];
            ++i;
        } else {
            nums[k] = R[j];
            ++j;
        }
        ++k;
    }
    
    // Copy the remaining elements of L[], if there are any
    while (i < n1) {
        nums[k] = L[i];
        ++i;
        ++k;
    }
    
    // Copy the remaining elements of R[], if there are any
    while (j < n2) {
        nums[k] = R[j];
        ++j;
        ++k;
    }
}

void mergeSort(std::vector<int>& nums, int left, int right) {
    if (left < right) {
        int mid = left + (right - left) / 2;
        
        // Sort first and second halves
        mergeSort(nums, left, mid);
        mergeSort(nums, mid + 1, right);
        
        // Merge the sorted halves
        merge(nums, left, mid, right);
    }
}

std::vector<int> sortArray(std::vector<int>& nums) {
    if (nums.empty()) {
        return nums;
    }
    mergeSort(nums, 0, nums.size() - 1);
    return nums;
}


// vector<int> sortArray(std::vector<int>& a) {
//     int n = a.size();
//     for(int j = 1;j<n;j++){
//         int key = a[j];
//         int i = j-1;
//         while(i>=0 && a[i]>key){
//             a[i+1] = a[i];
//             --i;
//         }
//         a[i+1] = key;
//     }
//     return a;
// }

};
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