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

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  • 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.

/* 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).

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