Class Solution

  • All Implemented Interfaces:

    
    public final class Solution
    
                        

    2423 - Remove Letter To Equalize Frequency.

    Easy

    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 chose to do nothing.

    Example 1:

    Input: word = "abcc"

    Output: true

    Explanation: Select index 3 and delete it: word becomes "abc" and each character has a frequency of 1.

    Example 2:

    Input: word = "aazz"

    Output: false

    Explanation: We must delete a character, so either the frequency of "a" is 1 and the frequency of "z" is 2, or vice versa. It is impossible to make all present letters have equal frequency.

    Constraints:

    • 2 <= word.length <= 100

    • word consists of lowercase English letters only.

    • Nested Class Summary

      Nested Classes 
      Modifier and Type Class Description
    • Field Summary

      Fields 
      Modifier and Type Field Description
    • Constructor Summary

      Constructors 
      Constructor Description
      Solution()
    • Enum Constant Summary

      Enum Constants 
      Enum Constant Description
    • Method Summary

      Modifier and Type Method Description
      final Boolean equalFrequency(String word)
      • Methods inherited from class java.lang.Object

        clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait