Is Remove Letter To Equalize Frequency easy or hard?

The problem is classified as Easy on LeetCode but can be tricky because you must reason about frequency patterns after a single removal. Once you focus on character counts instead of positions, the solution becomes straightforward with a simple validation check.

Remove Letter To Equalize Frequency Python/Java solution

Both Python and Java solutions typically use a frequency array or hash map to count characters. After building the counts, simulate removing one occurrence of each letter and verify whether the remaining frequencies are equal. This keeps the implementation simple while maintaining O(n) time complexity.

How to solve Remove Letter To Equalize Frequency in O(n)?

First count the frequency of each letter using an array or hash map. Then try decreasing the count of each character once and check if all remaining non-zero frequencies are equal. Because there are at most 26 characters, each validation step is constant time, giving an overall O(n) algorithm.

What is the best approach for Remove Letter To Equalize Frequency?

The most efficient approach uses a frequency map of the 26 lowercase letters and simulates removing one occurrence of each character type. After each simulated removal, check whether all remaining non-zero frequencies match. This runs in O(n) time with O(1) space because the alphabet size is constant.

Is Remove Letter To Equalize Frequency asked at Google/Amazon/Meta?

Frequency manipulation and counting problems like this appear frequently in interviews at companies such as Amazon, Google, and Meta. While this exact problem may vary, the underlying pattern—analyzing character counts with a hash table—is common in string interview questions.

What data structure is used in Remove Letter To Equalize Frequency?

The primary data structure is a hash table or a fixed-size integer array that stores character frequencies. Since the input consists of lowercase English letters, a 26-element array is usually the most efficient representation.

What is the time complexity of Remove Letter To Equalize Frequency?

The optimal solution runs in O(n) time where n is the length of the string. Building the frequency map requires one pass through the string, and validation checks at most 26 characters. Space complexity is O(1) since only a fixed-size frequency array is used.

Description Hints 2 Solution Video 10:35

2423. Remove Letter To Equalize Frequency

Easy19.3% acceptance4 companies

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: