Longest Common Substring

//M1
/*
We will take a dp array of n*m where 1 to n(rows) represents the index of string1 and same goes for m(columns)
arr[i][j] represents the length of substring ending at ith index in s1 and jtj index in s2 because it is
substring so at (i,j) we will be having answer for (i-1,j-1)th index.
*/
class Solution{
    public:

    int longestCommonSubstr (string text1, string text2, int n, int m){
        int n1=text1.size(),n2=text2.size();
        int ma=0;
          int arr[n1+1][n2+1];
            for(int i=0;i<=n1;i++)   //0 size string can form only 0 size common substring with any size substring
                arr[i][0]=0;
            for(int j=0;j<=n2;j++)   //0 size substring can form only 0 size common substring with any size string
                 arr[0][j]=0;
            for(int i=1;i<=n1;i++){   //if current character matches,then maximum length of substring upto that
                for(int j=1;j<=n2;j++){// will be the maximum of last visited character ie. arr[i-1][j-1]+1;
                    if(text1[i-1]==text2[j-1])
                         arr[i][j]=arr[i-1][j-1]+1;
                    else arr[i][j]=0;    //if not matches,then substring can't go ahead and current substring broke
                    ma=max(arr[i][j],ma); //so arr[i][j]=0;
                }
            }
        return ma;  //for finding max_length substring,we wen through whole dp table
    }
};
//M2
 public int lcsSubstrMemo(char[] s1, char[] s2, int m, int n, int c, int[][] t) {
            if(m == 0 || n == 0) {
                return c;
            }
            if (t[m-1][n-1] != -1) return t[m-1][n-1];
            if(s1[m - 1] == s2[n - 1]) {
                c = lcsSubstr(s1, s2, m - 1, n - 1, c + 1);
            } else {
                c2 = Math.max(lcsSubstr(s1, s2, m, n - 1, 0), lcsSubstr(s1, s2, m - 1, n, 0));
            }
            t[m - 1][n - 1] = Math.max(c, c2);
            return t[m-1][n-1];
        }