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Double Sequence X

Description

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

A 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.

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

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

Example I:

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

Example II:

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

Example III:

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

Questions

  • LeetCode - 1143. Longest Common Subsequence

Transform Function

for(int i = 1; i <= m; i++) {
    for(int j = 1; j <= n; j++) {
        if(c1[i - 1] == c2[j - 1]) {
            dp[i][j] = dp[i - 1][j - 1] + 1;
        }else{
            dp[i][j] = Math.max(dp[i - 1][j - 1], Math.max(dp[i - 1][j], dp[i][j - 1]));
        }
    }
}

Solution I

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        char[] c1 = text1.toCharArray();
        char[] c2 = text2.toCharArray();
        int m = c1.length, n = c2.length;
        int[][] dp = new int[m + 1][n + 1];
        for(int i = 1; i <= m; i++) {
            for(int j = 1; j <= n; j++) {
                if(c1[i - 1] == c2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }else{
                    dp[i][j] = Math.max(dp[i - 1][j - 1], Math.max(dp[i - 1][j], dp[i][j - 1]));
                }
            }
        }
        return dp[m][n];
    }
}