First Letter to Appear Twice

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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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

✓ Bit Manipulation

⏱️ Time: O(n) Space: O(1)

Since we only have 26 lowercase letters, we can use a 32-bit integer as a bitmask. Each bit represents whether we've seen a particular letter. When we encounter a letter, we check if its corresponding bit is set.

Brute Force

⏱️ Time: O(n²) Space: O(1)

Iterate through the string and for each character, scan backwards through all previous positions to check if this character has appeared before. Return the first character where we find a duplicate.

Hash Set

⏱️ Time: O(n) Space: O(1)

Iterate through the string once, maintaining a hash set of characters we've seen. When we encounter a character that's already in the set, we've found our answer.

Bit Manipulation — Algorithm Steps

Step 1: Initialize bitmask to 0
Step 2: For each character, calculate bit position
Step 3: If bit is set, return character; otherwise set the bit

Visualization

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Step-by-Step Walkthrough

Initialize

Start with bitmask = 0

Check Bit

For each char, check corresponding bit

Found

Bit already set means duplicate

Code -

solution.c — C

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

char solution(char* s) {
    int seen = 0;
    int len = strlen(s);
    for (int i = 0; i < len; i++) {
        int bit = 1 << (s[i] - 'a');
        if (seen & bit) {
            return s[i];
        }
        seen |= bit;
    }
    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(n)

Single pass through string, bit operations are O(1)

✓ Linear Growth

Space Complexity

O(1)

Only using one integer variable for bitmask

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Bit Manipulation — Algorithm Steps

Visualization

Code -

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