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Kotlin-based LeetCode algorithm problem solutions, regularly updated
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.