[NeetCode 150] Longest Common Subsequence

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Longest Common Subsequence

Given two strings text1 and text2, return the length of the longest common subsequence between the two strings if one exists, otherwise return 0.

A subsequence is a sequence that can be derived from the given sequence by deleting some or no elements without changing the relative order of the remaining characters.

For example, “cat” is a subsequence of “crabt”.
A common subsequence of two strings is a subsequence that exists in both strings.

Example 1:

Input: text1 = "cat", text2 = "crabt" 

Output: 3

Explanation: The longest common subsequence is “cat” which has a length of 3.

Example 2:

Input: text1 = "abcd", text2 = "abcd"

Output: 4

Example 3:

Input: text1 = "abcd", text2 = "efgh"

Output: 0

Constraints:

1 <= text1.length, text2.length <= 1000

text1 and text2 consist of only lowercase English characters.

Solution

Let $d p [i] [j]$ represents the length of longest common subsequence between $text1_{0:i}$ and $text2_{0:j}$ . If $t e x t 1 [i] == t e x t 2 [j]$ , then we can transfer from $d p [i - 1] [j - 1]$ that $d p [i] [j] = d p [i - 1] [j - 1] + 1$ . Or $d p [i] [j]$ will be $max⁡(dp[i−1][j],dp[i][j−1]])\max(dp[i-1][j], dp[i][j-1]])$ , because $t e x t 1 [i]$ and $t e x t 2 [j]$ do not make any contribution to the common subsequence.

Code

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        dp = [[0]*len(text2) for _ in range(len(text1))]
        for j in range(len(text2)):
            dp[0][j] = 1 if text1[0] == text2[j] else 0
        for j in range(1, len(text2)):
            dp[0][j] = max(dp[0][j], dp[0][j-1])
        for i in range(1, len(text1)):
            dp[i][0] = 1 if text1[i] == text2[0] else 0
            dp[i][0] = max(dp[i][0], dp[i-1][0])
            for j in range(1, len(text2)):
                if text1[i] == text2[j]:
                    dp[i][j] = dp[i-1][j-1]+1
                else:
                    dp[i][j] = max(dp[i-1][j], dp[i][j-1])
        print(dp)
        return dp[-1][-1]