Time complexity of insertion sort when there are O(n) inversions?

What is an inversion?
Given an array arr[], a pair arr[i] and arr[j] forms an inversion if arr[i] j. For example, the array {1, 3, 2, 5} has one inversion (3, 2) and array {5, 4, 3} has inversions (5, 4), (5, 3) and (4, 3). We have discussed a merge sort based algorithm to count inversions

What is the time complexity of Insertion Sort when there are O(n) inversions?
Consider the following function of insertion sort.

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/* Function to sort an array using insertion sort*/
void insertionSort(int arr[], int n)
{
   int i, key, j;
   for (i = 1; i < n; i++)
   {
       key = arr[i];
       j = i-1;
   
       /* Move elements of arr[0..i-1], that are
          greater than key, to one position ahead
          of their current position */
       while (j >= 0 && arr[j] > key)
       {
           arr[j+1] = arr[j];
           j = j-1;
       }
       arr[j+1] = key;
   }
}
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If we take a closer look at the insertion sort code, we can notice that every iteration of while loop reduces one inversion. The while loop executes only if i > j and arr[i] 2).

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