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* ci(pre-commit): Add pep8-naming to `pre-commit` hooks (#7038) * refactor: Fix naming conventions (#7038) * Update arithmetic_analysis/lu_decomposition.py Co-authored-by: Christian Clauss <cclauss@me.com> * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * refactor(lu_decomposition): Replace `NDArray` with `ArrayLike` (#7038) * chore: Fix naming conventions in doctests (#7038) * fix: Temporarily disable project euler problem 104 (#7069) * chore: Fix naming conventions in doctests (#7038) Co-authored-by: Christian Clauss <cclauss@me.com> Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
106 lines
2.9 KiB
Python
106 lines
2.9 KiB
Python
"""
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Author : Turfa Auliarachman
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Date : October 12, 2016
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This is a pure Python implementation of Dynamic Programming solution to the edit
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distance problem.
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The problem is :
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Given two strings A and B. Find the minimum number of operations to string B such that
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A = B. The permitted operations are removal, insertion, and substitution.
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"""
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class EditDistance:
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"""
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Use :
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solver = EditDistance()
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editDistanceResult = solver.solve(firstString, secondString)
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"""
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def __init__(self):
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self.__prepare__()
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def __prepare__(self, n=0, m=0):
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self.dp = [[-1 for y in range(0, m)] for x in range(0, n)]
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def __solve_dp(self, x, y):
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if x == -1:
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return y + 1
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elif y == -1:
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return x + 1
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elif self.dp[x][y] > -1:
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return self.dp[x][y]
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else:
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if self.a[x] == self.b[y]:
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self.dp[x][y] = self.__solve_dp(x - 1, y - 1)
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else:
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self.dp[x][y] = 1 + min(
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self.__solve_dp(x, y - 1),
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self.__solve_dp(x - 1, y),
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self.__solve_dp(x - 1, y - 1),
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)
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return self.dp[x][y]
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def solve(self, a, b):
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if isinstance(a, bytes):
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a = a.decode("ascii")
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if isinstance(b, bytes):
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b = b.decode("ascii")
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self.a = str(a)
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self.b = str(b)
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self.__prepare__(len(a), len(b))
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return self.__solve_dp(len(a) - 1, len(b) - 1)
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def min_distance_bottom_up(word1: str, word2: str) -> int:
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"""
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>>> min_distance_bottom_up("intention", "execution")
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5
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>>> min_distance_bottom_up("intention", "")
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9
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>>> min_distance_bottom_up("", "")
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0
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"""
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m = len(word1)
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n = len(word2)
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dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]
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for i in range(m + 1):
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for j in range(n + 1):
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if i == 0: # first string is empty
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dp[i][j] = j
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elif j == 0: # second string is empty
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dp[i][j] = i
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elif (
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word1[i - 1] == word2[j - 1]
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): # last character of both substing is equal
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dp[i][j] = dp[i - 1][j - 1]
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else:
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insert = dp[i][j - 1]
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delete = dp[i - 1][j]
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replace = dp[i - 1][j - 1]
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dp[i][j] = 1 + min(insert, delete, replace)
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return dp[m][n]
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if __name__ == "__main__":
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solver = EditDistance()
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print("****************** Testing Edit Distance DP Algorithm ******************")
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print()
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S1 = input("Enter the first string: ").strip()
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S2 = input("Enter the second string: ").strip()
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print()
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print("The minimum Edit Distance is: %d" % (solver.solve(S1, S2)))
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print("The minimum Edit Distance is: %d" % (min_distance_bottom_up(S1, S2)))
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print()
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print("*************** End of Testing Edit Distance DP Algorithm ***************")
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