**What is an inversion?**

Given an array arr[], a pair arr[i] and arr[j] forms an inversion if arr[i] < arr[j] and 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.

`/* 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;` ` ` `}` `}` |

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] < arr[j]. Therefore total number of while loop iterations (For all values of i) is same as number of inversions. Therefore overall time complexity of the insertion sort is O(n + f(n)) where f(n) is inversion count. If the inversion count is O(n), then the time complexity of insertion sort is O(n).
In worst case, there can be n*(n-1)/2 inversions. The worst case occurs when the array is sorted in reverse order. So the worst case time complexity of insertion sort is O(n^{2}).

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