LeetCode 1143. Longest Common Subsequence

Description

https://leetcode.com/problems/longest-common-subsequence/

Given two strings text1 and text2, return the length of their longest common subsequence.

subsequence of a string is a new string generated from the original string with some characters(can be none) deleted without changing the relative order of the remaining characters. (eg, “ace” is a subsequence of “abcde” while “aec” is not). A common subsequence of two strings is a subsequence that is common to both strings.

If there is no common subsequence, return 0.

Example 1:

Input: text1 = "abcde", text2 = "ace" 
Output: 3  
Explanation: The longest common subsequence is "ace" and its length is 3.

Example 2:

Input: text1 = "abc", text2 = "abc"
Output: 3
Explanation: The longest common subsequence is "abc" and its length is 3.

Example 3:

Input: text1 = "abc", text2 = "def"
Output: 0
Explanation: There is no such common subsequence, so the result is 0.

Constraints:

  • 1 <= text1.length <= 1000
  • 1 <= text2.length <= 1000
  • The input strings consist of lowercase English characters only.

Explanation

Try dynamic programming. DP[i][j] represents the longest common subsequence of text1[0 … i] & text2[0 … j].

DP[i][j] = DP[i – 1][j – 1] + 1 , if text1[i] == text2[j] DP[i][j] = max(DP[i – 1][j], DP[i][j – 1]) , otherwise

Python Solution

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        
        f = [[0 for j in range(0, len(text2) + 1)] for i in range(0, len(text1) + 1)]

    
        for i in range(1, len(text1) + 1):
            for j in range (1, len(text2) + 1):
                if text1[i - 1] == text2[j - 1]:
                    f[i][j] = f[i - 1][j - 1] + 1
                else:
                    f[i][j] = max(f[i - 1][j], f[i][j - 1])
                    
                    
                
        return f[len(text1)][len(text2)]
  • Time complexity: O(MN). We’re solving M * Nsubproblems. Solving each subproblem is an O(1)operation.
  • Space complexity: O(MN).We’re allocating a 2D array of size M * N to save the answers to subproblems.

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