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# 1044.Longest-Duplicate-Substring

## 题目描述

Given a string S, consider all duplicated substrings: (contiguous) substrings of S that occur 2 or more times. (The occurrences may overlap.)
Return any duplicated substring that has the longest possible length. (If S does not have a duplicated substring, the answer is "".)
Example 1:
Input: "banana"
Output: "ana"
Example 2:
Input: "abcd"
Output: ""
Note:
2 <= S.length <= 10^5
S consists of lowercase English letters.

## 代码

### Approach 1: Binary Search + Rabin-Karp

Complexity Analysis
• Time complexity : O(N_log_N). O(logN) for the binary search and O(N) for Rabin-Karp algorithm.
• Space complexity : O(N) to keep the hashset.
class Solution {
public String longestDupSubstring(String S) {
int n = S.length();
int[] nums = new int[n];
for (int i = 0; i < n; i++) {
nums[i] = (int)S.charAt(i) - (int)'a';
}
// base value for the rolling hash function
int a = 26;
// modulus value for the rolling hash function to avoid overflow
long modulus = (long)Math.pow(2, 32);
int left = 1, right = n;
int L;
while (left <= right) {
L = left + (right - left) / 2;
if (search(L, a, modulus, n, nums) == -1) {
right = L - 1;
} else {
left = L + 1;
}
}
int start = search(left - 1, a, modulus, n, nums);
return S.substring(start, start + left - 1);
}
public int search(int L, int a, long modulus, int n, int[] nums) {
// compute the hash of string S[:L]
long h = 0;
for (int i = 0; i < L; i++) {
h = (h * a + nums[i]) % modulus;
}
HashSet<Long> seen = new HashSet();
// const value to be used often : a**L % modulus
long aL = 1;
for (int i = 1; i <= L; i++) {
aL = (aL * a) % modulus;
}
for (int start = 1; start < n - L + 1; start++) {
// compute rolling hash in O(1) time
h = (h * a - nums[start - 1] * aL % modulus + modulus) % modulus;
h = (h + nums[start + L - 1]) % modulus;
if (seen.contains(h)) {
return start;
} else {