If the array of size n contains just one unique number (an array with all duplicates), then the traditional quick sort algorithm will create very uneven partitions of sizes 1 and n-1. This will be the case no matter how we choose the pivot element. This would make the average case time complexity O(n^2) for such inputs.
I thought of a simple way to alleviate this problem. While partitioning the array, if the element is exactly equal to the pivot element, we can equi-probably (randomly through a coin toss) choose to treat it as "less than" or "greater than" the pivot, so that duplicates get evenly distributed around the pivot.
def partition(arr: list[int]): pivot = arr[-1] current_index = 0 for i in range(len(arr) - 1): if arr[i] < pivot or (arr[i] == pivot and random.random() < 0.5): arr[i], arr[current_index] = arr[current_index], arr[i] current_index += 1 # put the pivot in its place arr[-1], arr[current_index] = arr[current_index], arr[-1] # return the partition index return current_index(Emphasis on the arr[i] == pivot and random.random() < 0.5 check)
But I have never seen this approach being used anywhere. Is there a problem with this approach that it's not widely used?
