# Longest Repeating Substring - Google Top Interview Questions

### Problem Statement :

Given a lowercase alphabet string s, return the length of the longest substring that occurs at least two times in s. If there's no such string, return 0.

Constraints

0 ≤ n ≤ 1,000 where n is the length of s

Example 1

Input

s = "abcdzabcd"

Output

4

Explanation

The longest substring that occurs more than once is "abcd".

Example 2

Input

s = "abcdefg"

Output

0

Explanation

There's no repeating substring.

### Solution :

Solution in C++ :

string s;
int n;

struct node {
map<char, int> next;

node(int l = 0, int r = 0, int par = -1) : l(l), r(r), par(par), link(-1) {
}
int len() {
return r - l;
}
int &get(char c) {
if (!next.count(c)) next[c] = -1;
return next[c];
}
};
node t[2005];
int sz;
int ans;

void dfs(int curr, int len) {
len += t[curr].r - t[curr].l;
if (t[curr].next.size()) {
ans = max(ans, len);
for (auto out : t[curr].next) {
dfs(out.second, len);
}
}
}

struct state {
int v, pos;
state(int v, int pos) : v(v), pos(pos) {
}
};
state ptr(0, 0);

state go(state st, int l, int r) {
while (l < r)
if (st.pos == t[st.v].len()) {
st = state(t[st.v].get(s[l]), 0);
if (st.v == -1) return st;
} else {
if (s[t[st.v].l + st.pos] != s[l]) return state(-1, -1);
if (r - l < t[st.v].len() - st.pos) return state(st.v, st.pos + r - l);
l += t[st.v].len() - st.pos;
st.pos = t[st.v].len();
}
return st;
}

int split(state st) {
if (st.pos == t[st.v].len()) return st.v;
if (st.pos == 0) return t[st.v].par;
node v = t[st.v];
int id = sz++;
t[id] = node(v.l, v.l + st.pos, v.par);
t[v.par].get(s[v.l]) = id;
t[id].get(s[v.l + st.pos]) = st.v;
t[st.v].par = id;
t[st.v].l += st.pos;
return id;
}

if (t[v].par == -1) return 0;
return t[v].link = split(go(state(to, t[to].len()), t[v].l + (t[v].par == 0), t[v].r));
}

void tree_extend(int pos) {
for (;;) {
state nptr = go(ptr, pos, pos + 1);
if (nptr.v != -1) {
ptr = nptr;
return;
}

int mid = split(ptr);
int leaf = sz++;
t[leaf] = node(pos, n, mid);
t[mid].get(s[pos]) = leaf;

ptr.pos = t[ptr.v].len();
if (!mid) break;
}
}

void build_tree() {
sz = 1;
ptr = state(0, 0);
for (int i = 0; i < n; ++i) tree_extend(i);
}

int solve(string str) {
str += "\$";
s = str;
n = str.size();
build_tree();
ans = 0;
dfs(0, 0);
for (int i = 0; i <= n; i++) {
t[i] = node();
}
return ans;
}

Solution in Java :

import java.util.*;

class Solution {
public class RabinKarp {
String s;
long[] prefix_hash;
long[] powers;

final int PRIME1 = 1009;
final int MOD1 = 998244353;

/**
* Class constructor specifiying the string s we want to work with.
*/
RabinKarp(String s) {
this.s = s;
this.prefix_hash = new long[this.s.length() + 1];
this.powers = new long[this.s.length() + 1];
this.computeHashAndPow();
}
/**
* Computes the prefix hash values and computes the prefix powers.
*/
private void computeHashAndPow() {
this.powers[0] = 1;
for (int i = 1; i <= this.s.length(); i++) {
long c = (long) this.s.charAt(i - 1) - 'a' + 1;
this.prefix_hash[i] = ((prefix_hash[i - 1] * PRIME1 + c) % MOD1);
this.powers[i] = ((this.powers[i - 1] * PRIME1) % MOD1);
}
}

/**
* Takes a left and right inclusive indices that resembles a substring and calculates the
* hash in O(1) time
* @param l The left of the substring
* @param r The right of the substring
* @return The hash of the substring [l,r].
*/
public long getHashSubstring(int l, int r) {
return (this.prefix_hash[r + 1] - this.prefix_hash[l] * this.powers[r - l + 1] % MOD1
+ MOD1)
% MOD1;
}
}
public int solve(String s) {
int n = s.length();
RabinKarp rk = new RabinKarp(s);
int res = 0;

int left = 1;
int right = n;

while (left <= right) {
int candidate_length = left + (right - left) / 2;
HashSet<Long> seen = new HashSet();

boolean worked = false;

for (int i = 0; i < n; i++) {
if (i + candidate_length - 1 < n) {
long hash = rk.getHashSubstring(i, i + candidate_length - 1);
if (seen.contains(hash)) {
worked = true;
res = Math.max(res, candidate_length);
}
}
}
if (worked) {
left = candidate_length + 1;
} else {
right = candidate_length - 1;
}
}
return res;
}
}

Solution in Python :

MOD = 10 ** 9 + 7
P = 53

class Solution:
def solve(self, s):
N = len(s)

def can(x):
if x <= 0:
return True
if x > N:
return False
hash = 0
power = (P ** (x - 1)) % MOD
hash_to_idx = defaultdict(list)  # maps hash => start index of substring

for i in range(x):
hash = (hash * P + ord(s[i])) % MOD
hash_to_idx[hash].append(0)
for i in range(x, N):
hash = (hash - power * ord(s[i - x])) % MOD
hash = (hash * P + ord(s[i])) % MOD
start_idx = i - x + 1
if hash in hash_to_idx:
this_string = s[start_idx : i + 1]
for other_idx in hash_to_idx[hash]:
other_string = s[other_idx : other_idx + x]
if this_string == other_string:
return True
hash_to_idx[hash].append(start_idx)
return False

lo = 0
hi = N + 1
while lo < hi - 1:
mid = (lo + hi) // 2
if can(mid):
lo = mid
else:
hi = mid
return lo

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