// Java implementation of
// finding length of longest
// Common substring using
// Dynamic Programming
class GFG {
	/*
	Returns length of longest common substring
	of X[0..m-1] and Y[0..n-1]
	*/
	static int LCSubStr(char X[], char Y[], int m, int n)
	{
		// Create a table to store
		// lengths of longest common
		// suffixes of substrings.
		// Note that LCSuff[i][j]
		// contains length of longest
		// common suffix of
		// X[0..i-1] and Y[0..j-1].
		// The first row and first
		// column entries have no
		// logical meaning, they are
		// used only for simplicity of program
		int LCStuff[][] = new int[m + 1][n + 1];
	
		// To store length of the longest
		// common substring
		int result = 0;

		// Following steps build
		// LCSuff[m+1][n+1] in bottom up fashion
		for (int i = 0; i <= m; i++)
		{
			for (int j = 0; j <= n; j++)
			{
				if (i == 0 || j == 0)
					LCStuff[i][j] = 0;
				else if (X[i - 1] == Y[j - 1])
				{
					LCStuff[i][j] = LCStuff[i - 1][j - 1] + 1;
					result = Integer.max(result, LCStuff[i][j]);
				}
				else
					LCStuff[i][j] = 0;
			}
		}
		return result;
	}

	// Driver Code
	public static void main(String[] args)
	{
		// Test 1:
		String X = "abcdabaabbacd";
		String Y = "abbabaabcabcd";
		int m = X.length();
		int n = Y.length();
		System.out.println(LCSubStr(X.toCharArray(), Y.toCharArray(), m, n));
		
		// Test 2:
		String X = "abcdefgh";
		String Y = "abaabcfg";
		int m = X.length();
		int n = Y.length();
		System.out.println(LCSubStr(X.toCharArray(), Y.toCharArray(), m, n));
	}
}
// Time Complexity -> O(m*n)
// where m and n are the lengths of the strings X and Y.