Time complexity of insertion sort when there are O(n) inversions? - GeeksforGeeks
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].Read full article from Time complexity of insertion sort when there are O(n) inversions? - GeeksforGeeks
