Largest Substring Between Two Equal Characters

Largest Substring Between Two Equal Characters - Problem

Given a string s, return the length of the longest substring between two equal characters, excluding the two characters. If there is no such substring return -1.

A substring is a contiguous sequence of characters within a string.

Input & Output

Example 1 — Basic Case

$ Input: s = "abaca"

› Output: 3

💡 Note: The longest substring between two equal characters is "bac" between the first 'a' (index 0) and last 'a' (index 4), which has length 3.

Example 2 — Multiple Pairs

$ Input: s = "aa"

› Output: 0

💡 Note: The two 'a' characters are adjacent with no characters between them, so the substring length is 0.

Example 3 — No Equal Characters

$ Input: s = "cbzxy"

› Output: -1

💡 Note: No two equal characters exist in the string, so return -1.

Constraints

1 ≤ s.length ≤ 300
s contains only lowercase English letters

Visualization

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Asked in

a Amazon 12 G Google 8

The key insight is that for each character, the maximum substring length is between its first and last occurrence. The optimal approach uses a single-pass hash map to track first occurrences and calculate lengths on the fly. Time: O(n), Space: O(k) where k is the number of unique characters.

Common Approaches

✓ Brute Force - Check All Pairs

⏱️ Time: O(n²) Space: O(1)

For each position in the string, check all positions after it to find matching characters. Calculate the substring length between each matching pair and keep track of the maximum.

Hash Map - Track First and Last Positions

⏱️ Time: O(n) Space: O(k)

Make two passes through the string. First pass stores the first occurrence of each character. Second pass updates with the last occurrence and calculates the maximum substring length.

Single Pass with Hash Map

⏱️ Time: O(n) Space: O(k)

Use a hash map to store the first occurrence of each character. As we iterate through the string, if we see a character again, calculate the substring length using its first occurrence and update the maximum.

Brute Force - Check All Pairs — Algorithm Steps

Step 1: For each character position i
Step 2: Check all positions j > i for matching character
Step 3: Calculate length j-i-1 and update maximum

Visualization

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Step-by-Step Walkthrough

Start at each position

Check character at position i

Compare with later positions

Find matching characters at positions j > i

Calculate length

Substring length = j - i - 1

Code -

solution.c — C

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

int solution(char* s) {
    int firstOccurrence[256];
    int maxLength = -1;
    int len = strlen(s);
    
    // Initialize array to -1 (indicating not found)
    for (int i = 0; i < 256; i++) {
        firstOccurrence[i] = -1;
    }
    
    // First pass: record first occurrence
    for (int i = 0; i < len; i++) {
        unsigned char c = (unsigned char)s[i];
        if (firstOccurrence[c] == -1) {
            firstOccurrence[c] = i;
        }
    }
    
    // Second pass: calculate max length
    for (int i = 0; i < len; i++) {
        unsigned char c = (unsigned char)s[i];
        if (firstOccurrence[c] != -1) {
            int length = i - firstOccurrence[c] - 1;
            if (length > maxLength) {
                maxLength = length;
            }
        }
    }
    
    return maxLength;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    
    // Remove newline if present
    int len = strlen(s);
    if (len > 0 && s[len-1] == '\n') {
        s[len-1] = '\0';
    }
    
    int result = solution(s);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops: outer loop n times, inner loop up to n times

✓ Linear Growth

Space Complexity

O(1)

Only using variables to track maximum length

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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