Count Odd Letters from Number

Count Odd Letters from Number - Problem

You are given an integer n. Perform the following steps:

Convert each digit of n into its lowercase English word (e.g., 4 → "four", 1 → "one")
Concatenate those words in the original digit order to form a string s
Return the number of distinct characters in s that appear an odd number of times

For example, if n = 234, we get "twothreefour", then count how many unique characters appear an odd number of times.

Input & Output

Example 1 — Basic Case

$ Input: n = 234

› Output: 6

💡 Note: 234 → "two" + "three" + "four" = "twothreefour". Character frequencies: t(2), w(1), o(2), h(1), r(3), e(3), f(1), u(1). Characters with odd frequency: w(1), h(1), r(3), e(3), f(1), u(1) = 6 total

Example 2 — Single Digit

$ Input: n = 5

› Output: 4

💡 Note: 5 → "five". Characters: f(1), i(1), v(1), e(1). All 4 characters appear once (odd) = 4 total

Example 3 — Repeated Digits

$ Input: n = 101

› Output: 1

💡 Note: 101 → "one" + "zero" + "one" = "onezeroone". Most characters appear even times, only 'z' appears 1 time (odd) = 1 total

Constraints

1 ≤ n ≤ 10⁹

Visualization

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Asked in

G Google 25 f Meta 18 a Amazon 15

The key insight is to convert each digit to its English word, concatenate them, then count character frequencies. The optimal approach uses a hash map for O(m) time complexity where m is the total string length. Time: O(m), Space: O(m).

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Dp 2d

⏱️ Time: N/A Space: N/A

Brute Force Character Counting

⏱️ Time: O(m²) Space: O(m)

First convert each digit to its English word and build the complete string. Then for each unique character, manually count how many times it appears in the string.

Hash Map Frequency Counting

⏱️ Time: O(m) Space: O(m)

Convert digits to words and concatenate them. Then use a hash map to count the frequency of each character in one pass. Finally count how many characters have odd frequencies.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

static int freq[256];

int solution(int n) {
    char* digitWords[] = {
        "zero", "one", "two", "three", "four",
        "five", "six", "seven", "eight", "nine"
    };
    
    // Clear frequency array
    memset(freq, 0, sizeof(freq));
    
    // Convert digits to words and count character frequencies
    char nStr[20];
    sprintf(nStr, "%d", n);
    
    for (int i = 0; nStr[i]; i++) {
        int digit = nStr[i] - '0';
        char* word = digitWords[digit];
        for (int j = 0; word[j]; j++) {
            freq[(unsigned char)word[j]]++;
        }
    }
    
    // Count characters with odd frequency
    int oddCount = 0;
    for (int i = 0; i < 256; i++) {
        if (freq[i] % 2 == 1) {
            oddCount++;
        }
    }
    
    return oddCount;
}

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

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

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