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50 lines
1.5 KiB
Plaintext
50 lines
1.5 KiB
Plaintext
---
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title: '1143. Longest Common Subsequence'
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description: Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0
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icon: dot
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topics:
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- { name: "String", slug: "string" }
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- { name: "Dynamic Programming", slug: "dynamic-programming" }
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- { name: "Longest Common Subsequence", slug: "longest-common-subsequence" }
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sidebar:
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label: 'Longest Common Subsequence'
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badge: 'Medium'
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---
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### Example 1:
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- Input: `text1 = "abcde", text2 = "ace"`
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- Output: `3`
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- Explanation: The longest common subsequence is "ace" and its length is `3`.
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### Example 2:
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- Input: `text1 = "abc", text2 = "abc"`
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- Output: `3`
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- Explanation: The longest common subsequence is "abc" and its length is `3`.
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### Example 3:
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- Input: `text1 = "abc", text2 = "def"`
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- Output: `0`
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- Explanation: There is no such common subsequence, so the result is `0`.
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### Constraints:
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- `1 <= text1.length, text2.length <= 1000`
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- `text1` and `text2` consist of only lowercase English characters.
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## Solution
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```py
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class Solution:
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def longestCommonSubsequence(self, text1: str, text2: str) -> int:
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dp = [[0 for j in range(len(text2) + 1)] for i in range(len(text1) + 1)]
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for i in range(len(text1) - 1, -1, -1):
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for j in range(len(text2) - 1, -1, -1):
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if text1[i] == text2[j]:
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dp[i][j] = 1 + dp[i + 1][j + 1]
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else:
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dp[i][j] = max(dp[i][j + 1], dp[i + 1][j])
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return dp[0][0]
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```
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