First Letter to Appear Twice - Practice Coding Problems

First Letter to Appear Twice - Problem

Given a string s consisting of lowercase English letters, return the first letter to appear twice.

Note:

A letter a appears twice before another letter b if the second occurrence of a is before the second occurrence of b.
s will contain at least one letter that appears twice.

Input & Output

Example 1 — Basic Case

$ Input: s = "abccba"

› Output: c

💡 Note: The first letter to appear twice is 'c' - it appears at positions 2 and 3

Example 2 — Early Duplicate

$ Input: s = "abba"

› Output: b

💡 Note: The first letter to appear twice is 'b' - it appears at positions 1 and 2

Example 3 — Immediate Repeat

$ Input: s = "aab"

› Output: a

💡 Note: The first letter to appear twice is 'a' - it appears at positions 0 and 1

Constraints

2 ≤ s.length ≤ 100
s consists of lowercase English letters
s has at least one repeating letter

Visualization

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Understanding the Visualization

Input

String with lowercase letters

Process

Track seen characters while iterating

Output

Return first duplicate found

Key Takeaway

🎯 Key Insight: Use a hash set to remember which characters you've seen - the first repeat is your answer

Asked in

M Meta 15 A Amazon 12

The key insight is to track which characters we've seen so far. Best approach is Hash Set with single pass through string. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O(n²)	O(1)	For each character, check all previous characters to see if it appeared before
Hash Set	O(n)	O(1)	Use a hash set to track seen characters in one pass
Bit Manipulation	O(n)	O(1)	Use a bitmask to track which characters we've seen

Brute Force — Algorithm Steps

Step 1: Iterate through each character in the string
Step 2: For each character, check all previous positions
Step 3: Return the first character that appears twice

Visualization

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Step-by-Step Walkthrough

Start

Begin with first character

Compare

For each char, check all previous

Found

Return first duplicate found

Code -

solution.c — C

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

char solution(char* s) {
    int len = strlen(s);
    for (int i = 0; i < len; i++) {
        for (int j = 0; j < i; j++) {
            if (s[i] == s[j]) {
                return s[i];
            }
        }
    }
    return '\0';
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops - for each character we scan all previous characters

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables, no extra data structures

✓ Linear Space

28.0K Views

Medium Frequency

~8 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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