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

• Difficulty Level : Easy
• Last Updated : 20 Dec, 2016

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.

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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;``   ``}``}`

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(n2).