Package g2401_2500.s2423_remove_letter_to_equalize_frequency

Class Solution

  • public class Solution
extends 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 Detail

      • Solution

        public Solution()

    • Method Detail

      • equalFrequency

        public boolean equalFrequency​(String word)