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ENH: refactored longest common subsequence, also fixed a bug with the sequence returned #1142
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5d81c55
function for the knapsack problem which returns one of the optimal su…
maxwell-aladago f2994cf
function for the knapsack problem which returns one of the optimal su…
maxwell-aladago 85b9c59
function for the knapsack problem which returns one of the optimal su…
maxwell-aladago 43fefa1
function for the knapsack problem which returns one of the optimal su…
maxwell-aladago 47f76eb
function for the knapsack problem which returns one of the optimal su…
maxwell-aladago 5e828fc
some pep8 cleanup too
maxwell-aladago eba05a9
ENH: refactored longest common subsequence, also fixed a bug with the…
maxwell-aladago ba9f6c3
Merge branch 'master' of https://github.com/TheAlgorithms/Python into…
maxwell-aladago 6c49319
renamed function
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Original file line number | Diff line number | Diff line change |
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""" | ||
LCS Problem Statement: Given two sequences, find the length of longest subsequence present in both of them. | ||
A subsequence is a sequence that appears in the same relative order, but not necessarily continious. | ||
A subsequence is a sequence that appears in the same relative order, but not necessarily continuous. | ||
Example:"abc", "abg" are subsequences of "abcdefgh". | ||
""" | ||
from __future__ import print_function | ||
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try: | ||
xrange # Python 2 | ||
except NameError: | ||
xrange = range # Python 3 | ||
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def lcs_dp(x, y): | ||
def lcs_dp(x: str, y: str): | ||
""" | ||
Finds the longest common subsequence between two strings. Also returns the | ||
The subsequence found | ||
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Parameters | ||
---------- | ||
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x: str, one of the strings | ||
y: str, the other string | ||
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Returns | ||
------- | ||
L[m][n]: int, the length of the longest subsequence. Also equal to len(seq) | ||
Seq: str, the subsequence found | ||
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>>> lcs_dp("programming", "gaming") | ||
(6, 'gaming') | ||
>>> lcs_dp("physics", "smartphone") | ||
(2, 'ph') | ||
>>> lcs_dp("computer", "food") | ||
(1, 'o') | ||
""" | ||
# find the length of strings | ||
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assert x is not None | ||
assert y is not None | ||
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m = len(x) | ||
n = len(y) | ||
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# declaring the array for storing the dp values | ||
L = [[None] * (n + 1) for i in xrange(m + 1)] | ||
seq = [] | ||
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for i in range(m + 1): | ||
for j in range(n + 1): | ||
if i == 0 or j == 0: | ||
L[i][j] = 0 | ||
elif x[i - 1] == y[ j - 1]: | ||
L[i][j] = L[i - 1][j - 1] + 1 | ||
seq.append(x[i -1]) | ||
L = [[0] * (n + 1) for _ in range(m + 1)] | ||
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for i in range(1, m + 1): | ||
for j in range(1, n + 1): | ||
if x[i-1] == y[j-1]: | ||
match = 1 | ||
else: | ||
L[i][j] = max(L[i - 1][j], L[i][j - 1]) | ||
# L[m][n] contains the length of LCS of X[0..n-1] & Y[0..m-1] | ||
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 = "" | ||
i, j = m, n | ||
while i > 0 and i > 0: | ||
if x[i - 1] == y[j - 1]: | ||
match = 1 | ||
else: | ||
match = 0 | ||
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if L[i][j] == L[i - 1][j - 1] + match: | ||
if match == 1: | ||
seq = x[i - 1] + seq | ||
i -= 1 | ||
j -= 1 | ||
elif L[i][j] == L[i - 1][j]: | ||
i -= 1 | ||
else: | ||
j -= 1 | ||
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return L[m][n], seq | ||
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if __name__=='__main__': | ||
x = 'AGGTAB' | ||
y = 'GXTXAYB' | ||
print(lcs_dp(x, y)) | ||
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if __name__ == '__main__': | ||
a = 'AGGTAB' | ||
b = 'GXTXAYB' | ||
expected_ln = 4 | ||
expected_subseq = "GTAB" | ||
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ln, subseq = lcs_dp(a, b) | ||
assert expected_ln == ln | ||
assert expected_subseq == subseq | ||
print("len =", ln, ", sub-sequence =", subseq) | ||
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Should we have a more intuitive function name?
I would just return the substring because the caller can easily do len(substring) if that information is really needed.