""" Given an array-like data structure A[1..n], how many pairs (i, j) for all 1 <= i < j <= n such that A[i] > A[j]? These pairs are called inversions. Counting the number of such inversions in an array-like object is the important. Among other things, counting inversions can help us determine how close a given array is to being sorted. In this implementation, I provide two algorithms, a divide-and-conquer algorithm which runs in nlogn and the brute-force n^2 algorithm. """ def count_inversions_bf(arr): """ Counts the number of inversions using a naive brute-force algorithm Parameters ---------- arr: arr: array-like, the list containing the items for which the number of inversions is desired. The elements of `arr` must be comparable. Returns ------- num_inversions: The total number of inversions in `arr` Examples --------- >>> count_inversions_bf([1, 4, 2, 4, 1]) 4 >>> count_inversions_bf([1, 1, 2, 4, 4]) 0 >>> count_inversions_bf([]) 0 """ num_inversions = 0 n = len(arr) for i in range(n - 1): for j in range(i + 1, n): if arr[i] > arr[j]: num_inversions += 1 return num_inversions def count_inversions_recursive(arr): """ Counts the number of inversions using a divide-and-conquer algorithm Parameters ----------- arr: array-like, the list containing the items for which the number of inversions is desired. The elements of `arr` must be comparable. Returns ------- C: a sorted copy of `arr`. num_inversions: int, the total number of inversions in 'arr' Examples -------- >>> count_inversions_recursive([1, 4, 2, 4, 1]) ([1, 1, 2, 4, 4], 4) >>> count_inversions_recursive([1, 1, 2, 4, 4]) ([1, 1, 2, 4, 4], 0) >>> count_inversions_recursive([]) ([], 0) """ if len(arr) <= 1: return arr, 0 mid = len(arr) // 2 P = arr[0:mid] Q = arr[mid:] A, inversion_p = count_inversions_recursive(P) B, inversions_q = count_inversions_recursive(Q) C, cross_inversions = _count_cross_inversions(A, B) num_inversions = inversion_p + inversions_q + cross_inversions return C, num_inversions def _count_cross_inversions(P, Q): """ Counts the inversions across two sorted arrays. And combine the two arrays into one sorted array For all 1<= i<=len(P) and for all 1 <= j <= len(Q), if P[i] > Q[j], then (i, j) is a cross inversion Parameters ---------- P: array-like, sorted in non-decreasing order Q: array-like, sorted in non-decreasing order Returns ------ R: array-like, a sorted array of the elements of `P` and `Q` num_inversion: int, the number of inversions across `P` and `Q` Examples -------- >>> _count_cross_inversions([1, 2, 3], [0, 2, 5]) ([0, 1, 2, 2, 3, 5], 4) >>> _count_cross_inversions([1, 2, 3], [3, 4, 5]) ([1, 2, 3, 3, 4, 5], 0) """ R = [] i = j = num_inversion = 0 while i < len(P) and j < len(Q): if P[i] > Q[j]: # if P[1] > Q[j], then P[k] > Q[k] for all i < k <= len(P) # These are all inversions. The claim emerges from the # property that P is sorted. num_inversion += len(P) - i R.append(Q[j]) j += 1 else: R.append(P[i]) i += 1 if i < len(P): R.extend(P[i:]) else: R.extend(Q[j:]) return R, num_inversion def main(): arr_1 = [10, 2, 1, 5, 5, 2, 11] # this arr has 8 inversions: # (10, 2), (10, 1), (10, 5), (10, 5), (10, 2), (2, 1), (5, 2), (5, 2) num_inversions_bf = count_inversions_bf(arr_1) _, num_inversions_recursive = count_inversions_recursive(arr_1) assert num_inversions_bf == num_inversions_recursive == 8 print("number of inversions = ", num_inversions_bf) # testing an array with zero inversion (a sorted arr_1) arr_1.sort() num_inversions_bf = count_inversions_bf(arr_1) _, num_inversions_recursive = count_inversions_recursive(arr_1) assert num_inversions_bf == num_inversions_recursive == 0 print("number of inversions = ", num_inversions_bf) # an empty list should also have zero inversions arr_1 = [] num_inversions_bf = count_inversions_bf(arr_1) _, num_inversions_recursive = count_inversions_recursive(arr_1) assert num_inversions_bf == num_inversions_recursive == 0 print("number of inversions = ", num_inversions_bf) if __name__ == "__main__": main()