Substrings That Begin and End With the Same Letter

Substrings That Begin and End With the Same Letter - Problem

Hash Table Math String Counting Prefix Sum Medium

You are given a 0-indexed string s consisting of only lowercase English letters. Return the number of substrings in s that begin and end with the same character.

A substring is a contiguous non-empty sequence of characters within a string.

Input & Output

Example 1 — Basic Case

$ Input: s = "abaca"

› Output: 8

💡 Note: Character 'a' appears 3 times: forms 3×4÷2=6 substrings. Characters 'b' and 'c' each appear once: each forms 1 substring. Total: 6+1+1=8

Example 2 — Single Character

$ Input: s = "aa"

› Output: 3

💡 Note: Character 'a' appears 2 times: forms 2×3÷2=3 substrings ("a", "a", "aa")

Example 3 — All Different

$ Input: s = "abc"

› Output: 3

💡 Note: Each character appears once: each forms 1 substring. Total: 1+1+1=3

Constraints

1 ≤ s.length ≤ 10⁵
s consists of only lowercase English letters

Visualization

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Asked in

G Google 25 a Amazon 18 M Microsoft 15

The key insight is that for n occurrences of any character, we can form exactly n×(n+1)/2 substrings that begin and end with that character. Best approach is frequency counting with mathematical formula. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check All Substrings

⏱️ Time: O(n³) Space: O(1)

Use nested loops to generate all possible substrings, then check if each substring's first and last characters are the same. This directly implements the problem definition.

Character Frequency Count

⏱️ Time: O(n) Space: O(1)

Count how many times each character appears, then for each character with count n, calculate n×(n+1)/2 which represents all possible substrings formed by that character.

Brute Force - Check All Substrings — Algorithm Steps

Generate all possible substrings using nested loops
For each substring, check if first character equals last character
Count valid substrings

Visualization

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Step-by-Step Walkthrough

Generate Substrings

Use nested loops to create all substrings

Check Endpoints

Compare first and last character of each substring

Count Valid

Increment counter for matching endpoints

Code -

solution.c — C

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

int solution(char* s) {
    int count = 0;
    int n = strlen(s);
    
    // Check all possible substrings
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            // Check if substring starts and ends with same character
            if (s[i] == s[j]) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;  // Remove newline
    
    int result = solution(s);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Outer loop O(n), inner loop O(n), substring comparison O(n)

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables, no extra data structures

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Ln 1, Col 1

Smart Actions

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Input & Output

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Related Problems

Common Approaches

Brute Force - Check All Substrings — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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