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 C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;

template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}
 
template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

const ll INF = 1e18;
const ll MOD = 998244353;
const ll MX = 1000001;

int N, ans;
string S;
int cnt[26];
vector<int> ocs[26];

void solve() {
    cin >> N >> S;
    
    for (int i = 0; i < N; i++) {
		ocs[S[i] - 97].push_back(i);
	} 
	for (int i = 0; i < N; i++) {
		int dec = S[i] - 97;
		// cout << dec << "\n";
		// cout << ocs[dec].size() << "\n";
		for (int oc = cnt[dec] + 1; oc < ocs[dec].size(); oc++) {
			int len = 0;
			int a = i, b = ocs[dec][oc];
			// cout << a << " " << b << "\n";
			while (a < ocs[dec][oc] && b < N && S[a] == S[b]) {
				len++;
				a++; b++;
			}
			
			chmax(ans, len);
		}
		cnt[dec]++;
	}
	cout << ans;
}

int main() {
	cin.sync_with_stdio(0); cin.tie(0);
    ll T = 1; // cin >> T;
    while (T--) solve();
    return 0;
    // You should actually read the stuff at the bottom
}

/* Stuff to Look For
 * -----------------
 * Int overflow, array bounds
 * Initializing all variables, avoid weird behavior
 * Edge cases(n = 0, n = 1)
 * Just return 0 after result
 */