- Difficulty: Easy
- Tags: LeetCode, Easy, Hash Table, String, Counting, leetcode-2423, O(n), O(1), Brute Force, Freq Table
Problem
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.
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.