Remove Letter To Equalize Frequency

1
2
3
4
5
6

class Solution:
    def equalFrequency(self, word: str) -> bool:
        for i in range(len(word)):
            if len(set(Counter(word[0:i] + word[i+1:]).values())) == 1:
                return True
        return False

This solution tries to solve the problem by attempting to remove each character from the string, one by one, and then checking if all remaining characters have the same frequency.

Here’s how it works:

  1. It iterates over the length of the word.

  2. For each iteration, it forms a new string by excluding the character at the current index (i). It does this by concatenating the part of the string before i (word[0:i]) with the part of the string after i (word[i+1:]).

  3. For each of these new strings, it computes the frequency of each character using the Counter function from the collections module. This function returns a dictionary where keys are characters and values are their counts in the string.

  4. Then it takes the values from the frequency dictionary, which represent counts of characters, and creates a set from them (set(Counter(...).values())). A set in Python is an unordered collection of unique elements, so if there were any duplicate counts in the original frequency dictionary, they will be removed in the set.

  5. The length of this set represents the number of different frequencies in the string. If all characters have the same frequency, the length of the set will be 1. If the length of the set is 1, it means that by removing the character at index i, we achieved a string where all characters have the same frequency. So the function returns True.

  6. If the function didn’t return True during the loop (which means it didn’t find a character that could be removed to equalize frequencies), it returns False after the loop.

This solution works because it exhaustively tries to remove every character to see if it results in equal frequencies. However, it’s not very efficient because it has to calculate frequencies for a new string in each iteration. The efficiency could be improved by calculating frequencies only once and then updating them during iterations.