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7592cba417
* added sol3.py for problem_20 * added sol4.py for problem_06 * ran `black .` on `\Python`
83 lines
2.0 KiB
Python
83 lines
2.0 KiB
Python
"""
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LCS Problem Statement: Given two sequences, find the length of longest subsequence present in both of them.
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A subsequence is a sequence that appears in the same relative order, but not necessarily continuous.
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Example:"abc", "abg" are subsequences of "abcdefgh".
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"""
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def longest_common_subsequence(x: str, y: str):
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"""
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Finds the longest common subsequence between two strings. Also returns the
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The subsequence found
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Parameters
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----------
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x: str, one of the strings
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y: str, the other string
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Returns
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-------
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L[m][n]: int, the length of the longest subsequence. Also equal to len(seq)
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Seq: str, the subsequence found
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>>> longest_common_subsequence("programming", "gaming")
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(6, 'gaming')
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>>> longest_common_subsequence("physics", "smartphone")
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(2, 'ph')
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>>> longest_common_subsequence("computer", "food")
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(1, 'o')
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"""
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# find the length of strings
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assert x is not None
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assert y is not None
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m = len(x)
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n = len(y)
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# declaring the array for storing the dp values
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L = [[0] * (n + 1) for _ in range(m + 1)]
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for i in range(1, m + 1):
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for j in range(1, n + 1):
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if x[i - 1] == y[j - 1]:
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match = 1
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else:
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match = 0
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L[i][j] = max(L[i - 1][j], L[i][j - 1], L[i - 1][j - 1] + match)
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seq = ""
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i, j = m, n
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while i > 0 and j > 0:
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if x[i - 1] == y[j - 1]:
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match = 1
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else:
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match = 0
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if L[i][j] == L[i - 1][j - 1] + match:
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if match == 1:
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seq = x[i - 1] + seq
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i -= 1
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j -= 1
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elif L[i][j] == L[i - 1][j]:
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i -= 1
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else:
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j -= 1
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return L[m][n], seq
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if __name__ == "__main__":
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a = "AGGTAB"
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b = "GXTXAYB"
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expected_ln = 4
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expected_subseq = "GTAB"
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ln, subseq = longest_common_subsequence(a, b)
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## print("len =", ln, ", sub-sequence =", subseq)
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import doctest
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doctest.testmod()
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