Remove letter to equalize frequency

Hash Table String Counting

19.2% acceptance

Feb 25, 2026

793

1353

You are given a 0-indexed string word, consisting of lowercase English letters. You need to select one index and remove the letter at that index from word so that the frequency of every letter present in word is equal. Return true if it is possible to remove one letter so that the frequency of all letters in word are equal, and false otherwise. Note: The frequency of a letter x is the number of times it occurs in the string. You must remove exactly one letter and cannot choose to do nothing.

Solution

Rust

Time O(n²)

Space O(1)

LeetCode

solution.rs

impl Solution {
  pub fn equal_frequency(word: String) -> bool {
    let n = word.len();
    let bytes = word.as_bytes();
    for i in 0..n {
      let mut freq = [0i32; 26];
      for j in 0..n {
        if j != i {
          freq[(bytes[j] - b'a') as usize] += 1;
        }
      }
      let nonzero: Vec<i32> = freq.iter().filter(|&&v| v > 0).cloned().collect();
      if nonzero.is_empty() {
        return true;
      }
      let first = nonzero[0];
      if nonzero.iter().all(|&v| v == first) {
        return true;
      }
    }
    false
  }
}