# 1044.Longest-Duplicate-Substring

## 1044. Longest Duplicate Substring

## 题目地址

<https://leetcode.com/problems/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(log*N*) for the binary search and O(*N*) for Rabin-Karp algorithm.
* Space complexity : O(*N*) to keep the hashset.

```java
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();
    seen.add(h);
    // 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 {
        seen.add(h);
      }
    }

    return -1;
  }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://wentao-shao.gitbook.io/leetcode/binary-search/1044.longest-duplicate-substring.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.