Distinct Echo Substrings

Distinct Echo Substrings - Problem

Given a string text, return the number of distinct non-empty substrings that can be written as the concatenation of some string with itself.

In other words, count all substrings that can be written as a + a where a is some non-empty string.

For example, if text = "abcabc", then "abcabc" can be written as "abc" + "abc", so it counts as an echo substring.

Input & Output

Example 1 — Basic Case

$ Input: text = "abab"

› Output: 1

💡 Note: Only "abab" is an echo substring ("ab" + "ab"). The substring "ab" by itself is not an echo since "a" ≠ "b".

Example 2 — Longer String

$ Input: text = "aabaaba"

› Output: 2

💡 Note: Echo substrings: "aa" appears at multiple positions but counts once, and "aabaab" ("aab" + "aab").

Example 3 — No Echoes

$ Input: text = "a"

› Output: 0

💡 Note: Single character cannot form an echo substring (need at least length 2)

Constraints

1 ≤ text.length ≤ 2000
text consists of lowercase English letters only

Visualization

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Asked in

G Google 15 F Facebook 12

The key insight is that an echo substring must have even length where the first half exactly matches the second half. The brute force approach checks all substrings in O(n³), while rolling hash optimizes comparisons to O(n²). Best approach uses rolling hash for efficient substring comparison. Time: O(n²), Space: O(n²)

Common Approaches

✓ Two Pointers

⏱️ Time: N/A Space: N/A

Brute Force

⏱️ Time: O(n³) Space: O(n³)

Generate all possible substrings, filter those with even length, then check if the first half matches the second half exactly. Use a set to track distinct echo substrings.

Rolling Hash Optimization

⏱️ Time: O(n²) Space: O(n²)

Instead of extracting substrings and comparing them directly, compute rolling hashes for the first and second halves of each potential echo substring. This eliminates expensive string operations.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

void solution(int* arr, int size) {
    // First pass: count how many zeros can be duplicated
    int duplicates = 0;
    for (int i = 0; i < size; i++) {
        if (arr[i] == 0) {
            // Check if we have space to duplicate this zero
            if (i + duplicates < size - 1) {
                duplicates++;
            } else if (i + duplicates == size - 1) {
                // Special case: zero at the last writable position
                // We can't duplicate it, but we need to handle it
                arr[size - 1] = 0;
                duplicates--;
                break;
            }
        }
    }
    
    // Second pass: place elements from right to left
    int writeIdx = size - 1;
    int readIdx = size - 1 - duplicates;
    
    while (readIdx >= 0 && writeIdx >= 0) {
        if (arr[readIdx] != 0) {
            arr[writeIdx] = arr[readIdx];
            writeIdx--;
        } else {
            // Duplicate zero
            arr[writeIdx] = 0;
            writeIdx--;
            if (writeIdx >= 0) {
                arr[writeIdx] = 0;
                writeIdx--;
            }
        }
        readIdx--;
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    // Parse array
    int arr[1000];
    int size = 0;
    
    // Handle empty array case
    if (strcmp(line, "[]") == 0) {
        printf("[]\n");
        return 0;
    }
    
    char* token = strtok(line + 1, ",]"); // Skip opening bracket
    while (token != NULL) {
        arr[size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    solution(arr, size);
    
    // Print result
    printf("[");
    for (int i = 0; i < size; i++) {
        printf("%d", arr[i]);
        if (i < size - 1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

⚠ Quadratic Space

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Medium Frequency

~35 min Avg. Time

847 Likes

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