Problem:
Score : 500 points

Problem Statement

Given is a string S of length N.

Find the maximum length of a non-empty string that occurs twice or more in S as contiguous substrings without overlapping.

More formally, find the maximum positive integer len such that there exist integers l_1 and l_2 ( 1 \leq l_1, l_2 \leq N - len + 1 ) that satisfy the following:

l_1 + len \leq l_2

S[l_1+i] = S[l_2+i] (i = 0, 1, ..., len - 1)

If there is no such integer len, print 0.

Constraints

2 \leq N \leq 5 \times 10^3

|S| = N

S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the maximum length of a non-empty string that occurs twice or more in S as contiguous substrings without overlapping. If there is no such non-empty string, print 0 instead.

Sample Input 1

5
ababa

Sample Output 1

2

The strings satisfying the conditions are: a, b, ab, and ba. The maximum length among them is 2, which is the answer.
Note that aba occurs twice in S as contiguous substrings, but there is no pair of integers l_1 and l_2 mentioned in the statement such that l_1 + len \leq l_2.

Sample Input 2

2
xy

Sample Output 2

0

No non-empty string satisfies the conditions.

Sample Input 3

13
strangeorange

Sample Output 3

5

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)



import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		String s = sc.next();
		int[][] dp = new int[n][n];


		for(int j = n-1; j>=0; j--) {
			int i = n-1;
			if(s.charAt(i) == s.charAt(j)) dp[i][j] = 1;
			else dp[i][j] = 0;
		}

		for(int i=n-1; i>=0; i--) {
			int j = n-1;
			if(s.charAt(i) == s.charAt(j)) dp[i][j] = 1;
			else dp[i][j] = 0;
		}

		for(int i = n-2; i>=0; i--) {
			for(int j = n-2; j>=0; j--) {
				if(s.charAt(i) == s.charAt(j)) dp[i][j] = dp[i+1][j+1]+1;
				else dp[i][j] = 0;
			}
		}

		int len = 0;
		for(int i=0; i<n; i++) {
			for(int j = 0; j<n; j++) {
				if(i+dp[i][j] <= j) len = Math.max(len, dp[i][j]);
			}
		}
		System.out.println(len);
	}
}