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Improve longest_common_substring.py (#12705)

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* Update longest_common_substring.py

- Combined the ans_index and ans_length into a single tuple to track the best match (position + length) more cleanly.

- Early exit for empty strings.

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* Update longest_common_substring.py

* [pre-commit.ci] auto fixes from pre-commit.com hooks

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Alfredo Hernandez Baeza 2025-05-10 07:57:43 -04:00 committed by GitHub
parent 59c3c8bbf3
commit 47a44abe23
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1 changed files with 9 additions and 6 deletions
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dynamic_programming/longest_common_substring.py
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@ -43,22 +43,25 @@ def longest_common_substring(text1: str, text2: str) -> str:
if not (isinstance(text1, str) and isinstance(text2, str)): if not (isinstance(text1, str) and isinstance(text2, str)):
raise ValueError("longest_common_substring() takes two strings for inputs") raise ValueError("longest_common_substring() takes two strings for inputs")
if not text1 or not text2:
return ""
text1_length = len(text1) text1_length = len(text1)
text2_length = len(text2) text2_length = len(text2)
dp = [[0] * (text2_length + 1) for _ in range(text1_length + 1)] dp = [[0] * (text2_length + 1) for _ in range(text1_length + 1)]
ans_index = 0 end_pos = 0
ans_length = 0 max_length = 0
for i in range(1, text1_length + 1): for i in range(1, text1_length + 1):
for j in range(1, text2_length + 1): for j in range(1, text2_length + 1):
if text1[i - 1] == text2[j - 1]: if text1[i - 1] == text2[j - 1]:
dp[i][j] = 1 + dp[i - 1][j - 1] dp[i][j] = 1 + dp[i - 1][j - 1]
if dp[i][j] > ans_length: if dp[i][j] > max_length:
ans_index = i end_pos = i
ans_length = dp[i][j] max_length = dp[i][j]
return text1[ans_index - ans_length : ans_index] return text1[end_pos - max_length : end_pos]
if __name__ == "__main__": if __name__ == "__main__":