Longest Duplicate Substring

Longest Duplicate Substring - Problem

String Binary Search Sliding Window Rolling Hash Suffix Array Hash Function Hard

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 "".

Input & Output

Example 1 — Basic Case

$ Input: s = "banana"

› Output: "ana"

💡 Note: "ana" appears twice in "banana" at positions 1-3 and 3-5. It's the longest substring that appears more than once.

Example 2 — No Duplicates

$ Input: s = "abcd"

› Output: ""

💡 Note: No substring appears more than once, so return empty string.

Example 3 — Single Character

$ Input: s = "aa"

› Output: "a"

💡 Note: Character 'a' appears twice, making it the longest (and only) duplicate substring.

Constraints

2 ≤ s.length ≤ 3 × 10⁴
s consists of lowercase English letters.

Visualization

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Asked in

G Google 25 f Facebook 20 a Amazon 15 M Microsoft 12

The key insight is using binary search on the answer length combined with polynomial rolling hash for efficient duplicate detection. The optimal approach achieves O(n² log n) time complexity by searching for the maximum length that has duplicates.

Common Approaches

✓ Prefix Sum

⏱️ Time: N/A Space: N/A

Greedy

⏱️ Time: N/A Space: N/A

Binary Search + Rolling Hash

⏱️ Time: O(n² log n) Space: O(n)

Use binary search to find the maximum length that has duplicates. For each length, use polynomial rolling hash to efficiently check if any substring of that length appears more than once.

Brute Force

⏱️ Time: O(n⁴) Space: O(n³)

Generate all possible substrings and check if each appears more than once. Start from longest possible length and work down to find the longest duplicate.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int countNegativeSubarrays(int* arr, int n) {
    long long* prefix = malloc((n + 1) * sizeof(long long));
    prefix[0] = 0;
    for (int i = 0; i < n; i++) {
        prefix[i + 1] = prefix[i] + arr[i];
    }
    
    int count = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 3; j <= n; j++) {
            if (prefix[j] - prefix[i] <= 0) {
                count++;
            }
        }
    }
    
    free(prefix);
    return count;
}

int canFixWithOperations(int* nums, int n, int remaining, int start, int* selected, int selectedLen) {
    if (remaining == 0) {
        int* temp = malloc(n * sizeof(int));
        memcpy(temp, nums, n * sizeof(int));
        for (int i = 0; i < selectedLen; i++) {
            temp[selected[i]] = 1000000000;
        }
        int result = countNegativeSubarrays(temp, n) == 0;
        free(temp);
        return result;
    }
    
    for (int i = start; i <= n - remaining; i++) {
        selected[selectedLen] = i;
        if (canFixWithOperations(nums, n, remaining - 1, i + 1, selected, selectedLen + 1)) {
            return 1;
        }
    }
    return 0;
}

int solution(int* nums, int n) {
    if (n <= 2) return 0;
    
    if (countNegativeSubarrays(nums, n) == 0) return 0;
    
    int* selected = malloc(n * sizeof(int));
    
    for (int k = 1; k <= n; k++) {
        if (canFixWithOperations(nums, n, k, 0, selected, 0)) {
            free(selected);
            return k;
        }
    }
    
    free(selected);
    return n;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int n = 0;
    char* token = strtok(line, "[],\n ");
    while (token) {
        nums[n++] = atoi(token);
        token = strtok(NULL, "[],\n ");
    }
    
    printf("%d\n", solution(nums, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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