Introduction

In this blog post, we will explore how to solve the LeetCode problem "2423" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's a breakdown of the solution:

• Frequency Analysis: Calculate the frequency of each character in the input string.
• Iteration and Removal: Iterate through the string, hypothetically removing each character one at a time. Recalculate frequencies after each removal.

• Equality Check: After each hypothetical removal, check if all the remaining characters have the same frequency.

• Runtime Complexity: O(n*26), where n is the length of the word. Storage Complexity: O(26) - constant space for the frequency map.

Code

    def equalFrequency(word: str) -> bool:
    """
    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.
    """

    def check_equal_frequency(freq):
        counts = [f for f in freq.values() if f > 0]
        if not counts:
            return True
        return len(set(counts)) == 1

    for i in range(len(word)):
        temp_word = word[:i] + word[i+1:]
        freq = {}
        for char in temp_word:
            freq[char] = freq.get(char, 0) + 1

        if check_equal_frequency(freq):
            return True

    return False