1143. Longest Common Subsequence

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        return helper(text1,text2,0,0);
    }
    
    int helper(String s1,String s2,int index1,int index2){
        if(index1==s1.length() || index2== s2.length()){
            return 0;
        }
        
        if(s1.charAt(index1) == s2.charAt(index2)){
            return 1 + helper(s1,s2,index1+1,index2+1);
        }
        
        return Math.max(helper(s1,s2,index1+1,index2),helper(s1,s2,index1,index2+1));
    }
}

class Solution {
    int[][] dp;
    public int longestCommonSubsequence(String text1, String text2) {
        dp = new int[text1.length()][text2.length()];
        for(int i=0;i<text1.length();i++){
            for(int j=0;j<text2.length();j++){
                dp[i][j]=-1;
            }
        }
        return helper(text1,text2,0,0);
    }
    
    int helper(String s1,String s2,int index1,int index2){
        if(index1==s1.length() || index2== s2.length()){
            return 0;
        }
        
        if(dp[index1][index2] >=0){
            return dp[index1][index2];
        }
        
        if(s1.charAt(index1) == s2.charAt(index2)){
            dp[index1][index2]= 1 + helper(s1,s2,index1+1,index2+1);
        }
        else{
            dp[index1][index2] = Math.max(helper(s1,s2,index1+1,index2),helper(s1,s2,index1,index2+1));
        }
        
        return dp[index1][index2];
    }
}

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