Longest Common Substring LeetCode Dynamic Programming

1143. Longest Common Subsequence

Medium

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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.

Accepted

105,172

Submissions

180,347

 

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        
        if(text1==null||text1.length()==0||text2 ==null||text2.length()==0){
            
            return 0;
        }
        
        int n = text1.length();
        
        int m = text2.length();
        
       int  dp[][] =  new  int[m+1][n+1];
        
        for(int i = 0; i<m; i++){
            
            dp[i][0]= 0;
        }
        
        for(int j = 0; j<n; j++){
            dp[0][j]=0;
        }
        
        int max = 0;
        for(int i = 1; i<=m; i++){
            for(int j = 1; j<=n;j++ ){
                
                
                if(text1.charAt(j-1)==text2.charAt(i-1)){
                    
                    dp[i][j]=dp[i-1][j-1]+1;
                }
                else{
                    
                   dp[i][j] = Math.max(dp[i][j-1],dp[i-1][j]); 
                }
  
            }
        }
        
        return dp[m][n];
    }
}