java.lang.Object
g2401_2500.s2423_remove_letter_to_equalize_frequency.Solution

public class Solution extends java.lang.Object
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.
  • Constructor Summary

    Constructors
    Constructor
    Description
     
  • Method Summary

    Modifier and Type
    Method
    Description
    boolean
    equalFrequency(java.lang.String word)
     

    Methods inherited from class java.lang.Object

    clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
  • Constructor Details

    • Solution

      public Solution()
  • Method Details

    • equalFrequency

      public boolean equalFrequency(java.lang.String word)