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