Count odd letters from number

Hash Table String Simulation Counting

85.3% acceptance

Mar 31, 2026

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.

Solution

Rust

Time O(n²)

Space O(1)

LeetCode

solution.rs

impl Solution {
  pub fn count_odd_letters(n: i32) -> i32 {
    let words = [
      "zero", "one", "two", "three", "four",
      "five", "six", "seven", "eight", "nine",
    ];
    
    let mut freq = [0u32; 26];
    let mut num = n;
    
    while num > 0 {
      let digit = (num % 10) as usize;
      for ch in words[digit].bytes() {
        freq[(ch - b'a') as usize] += 1;
      }
      num /= 10;
    }
    
    // Handle n == 0 edge case (but constraint says n >= 1)
    if n == 0 {
      for ch in words[0].bytes() {
        freq[(ch - b'a') as usize] += 1;
      }
    }
    
    freq.iter().filter(|&&c| c > 0 && c % 2 == 1).count() as i32
  }
}