1143. Longest Common Subsequence

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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.

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.

Solution

How to define the DP array?

dp[i][j] , denotes the LCS length of the first i chars in S1 and the first j chars in S2

 public int longestCommonSubsequence(String text1, String text2) {
        char[] s1 = text1.toCharArray();
        char[] s2 = text2.toCharArray();
        int N = s1.length;
        int M = s2.length;
        int[][] dp = new int[N+1][M+1];
        //dp[i][j] denotes the lcs of s1[0]~s1[i-1] and s2[0]~s2[j-1]
        for(int i = 0; i <= N; i++){
            for(int j = 0; j <= M; j++){
                if(i == 0 || j == 0){
                    dp[i][j] = 0;
                }
                else if(s1[i-1] == s2[j-1]){
                    dp[i][j] = dp[i-1][j-1] + 1;
                }
                else{
                    dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
                }
            }
        }
        return dp[N][M];
        
    }