Generate a String With Characters That Have Odd Counts

Generate a String With Characters That Have Odd Counts - Problem

String Easy

Given an integer n, return a string with n characters such that each character in the string occurs an odd number of times.

The returned string must contain only lowercase English letters. If there are multiple valid strings, return any of them.

Input & Output

Example 1 — Odd Length

$ Input: n = 3

› Output: aaa

💡 Note: String has 3 characters, character 'a' appears 3 times (odd count). Any single character repeated 3 times works.

Example 2 — Even Length

$ Input: n = 4

› Output: aaab

💡 Note: String has 4 characters, 'a' appears 3 times (odd), 'b' appears 1 time (odd). Both characters have odd counts.

Example 3 — Minimum Case

$ Input: n = 1

› Output: a

💡 Note: Single character string, 'a' appears 1 time (odd count). This is the smallest valid case.

Constraints

1 ≤ n ≤ 500

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is understanding parity: if n is odd, repeat one character n times (odd count). If n is even, use (n-1) of one character and 1 of another (both odd counts). Best approach is direct pattern. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Try Different Combinations

⏱️ Time: O(n²) Space: O(1)

Try different patterns of characters, count their frequencies, and verify that all characters appear an odd number of times. This is inefficient but demonstrates the basic logic.

Optimized - Direct Pattern

⏱️ Time: O(n) Space: O(1)

Skip the trial-and-error approach. If n is odd, use one character n times. If n is even, use one character (n-1) times and another character once. This ensures all characters appear an odd number of times.

Brute Force - Try Different Combinations — Algorithm Steps

Generate different character combinations
Count frequency of each character
Check if all frequencies are odd

Visualization

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Step-by-Step Walkthrough

Check n

Determine if n is odd or even

Apply Pattern

Use appropriate pattern based on n

Result

Return string with odd character counts

Code -

solution.c — C

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

char* solution(int n) {
    char* result = (char*)malloc((n + 1) * sizeof(char));
    // Pattern 1: all same character if n is odd
    if (n % 2 == 1) {
        for (int i = 0; i < n; i++) {
            result[i] = 'a';
        }
    }
    // Pattern 2: (n-1) of one char + 1 of another if n is even
    else {
        for (int i = 0; i < n - 1; i++) {
            result[i] = 'a';
        }
        result[n - 1] = 'b';
    }
    result[n] = '\0';
    return result;
}

int main() {
    int n;
    scanf("%d", &n);
    char* result = solution(n);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Generating combinations and counting frequencies for each attempt

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for counters

✓ Linear Space

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Medium Frequency

~10 min Avg. Time

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try Different Combinations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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