Count Odd Letters from Number - Practice Coding Problems

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: 4

💡 Note: 234 → "two" + "three" + "four" = "twothreefour". Characters with odd frequency: w(1), r(3), f(1), u(1) = 4 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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Understanding the Visualization

Input

Integer n = 234

Convert

2→two, 3→three, 4→four → twothreefour

Count

Find characters with odd frequencies

Key Takeaway

🎯 Key Insight: Convert digits to English words, concatenate, then use frequency counting to find characters appearing odd times

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

Approach	Time	Space	Notes
✓ Hash Map Frequency Counting	O(m)	O(m)	Use hash map to count character frequencies in a single pass
Brute Force Character Counting	O(m²)	O(m)	Convert digits to words, concatenate, then count each character frequency manually

Hash Map Frequency Counting — Algorithm Steps

Convert each digit to English word
Concatenate all words
Use hash map to count character frequencies
Count characters with odd frequency

Visualization

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Step-by-Step Walkthrough

Build String

Convert digits to words and concatenate

Count Frequencies

Single pass through string with hash map

Count Odds

Count characters with odd frequency

Code -

solution.c — C

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

int solution(int n) {
    char* digitWords[] = {"zero", "one", "two", "three", "four",
                         "five", "six", "seven", "eight", "nine"};
    
    // Convert digits to words and concatenate
    char nStr[20], s[1000] = "";
    sprintf(nStr, "%d", n);
    
    for (int i = 0; nStr[i]; i++) {
        strcat(s, digitWords[nStr[i] - '0']);
    }
    
    // Count character frequencies
    int freq[256] = {0};
    int len = strlen(s);
    for (int i = 0; i < len; i++) {
        freq[(unsigned char)s[i]]++;
    }
    
    // Count characters with odd frequency
    int oddCount = 0;
    for (int i = 0; i < 256; i++) {
        if (freq[i] > 0 && 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

⏱️

O(m)

Single pass through the concatenated string of length m

✓ Linear Growth

Space Complexity

O(m)

Hash map stores at most m distinct characters plus string storage

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Hash Map Frequency Counting — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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