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.
/* 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] 2).