How do you solve Remove Letter To Equalize Frequency on LeetCode?

Remove Letter To Equalize Frequency (LeetCode #2423) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding character frequency is crucial to solving this problem.

What is the brute force solution for Remove Letter To Equalize Frequency?

The brute force approach for Remove Letter To Equalize Frequency is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves removing each character from the string one by one and checking if the remaining characters have equal frequencies. This is straightforward but inefficient, as it requires multiple passes through the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Remove Letter To Equalize Frequency?

Remove Letter To Equalize Frequency uses the following concepts: Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Remove Letter To Equalize Frequency?

Always explain your thought process while coding. Consider edge cases and test your solution with them. Be prepared to discuss the time and space complexity of your solution.

What are common mistakes when solving Remove Letter To Equalize Frequency?

Not considering edge cases where all characters are the same. Failing to properly analyze frequency counts.

#2423

Remove Letter To Equalize Frequency

Easy

Hash TableStringCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach involves counting the frequency of each character and analyzing the frequency counts. By checking specific conditions on the frequency counts, we can determine if removing one character can equalize the frequencies efficiently.

⚙️

Algorithm

4 steps

1Step 1: Count the frequency of each character in the string.
2Step 2: Count how many times each frequency occurs.
3Step 3: Analyze the frequency counts to check if we can equalize them by removing one character.
4Step 4: If there are two distinct frequencies, check if one can be reduced to match the other by removing one character.

solution.py17 lines

1# Full working Python code
2
3def equalFrequency(word):
4    freq = {}
5    for char in word:
6        freq[char] = freq.get(char, 0) + 1
7    count = {}
8    for f in freq.values():
9        count[f] = count.get(f, 0) + 1
10    if len(count) == 1:
11        return True
12    if len(count) == 2:
13        keys = list(count.keys())
14        if (count[keys[0]] == 1 and (keys[0] - 1 == keys[1] or keys[0] == 1)) or 
15           (count[keys[1]] == 1 and (keys[1] - 1 == keys[0] or keys[1] == 1))):
16            return True
17    return False

ℹ

Complexity note: The time complexity is O(n) because we traverse the string to count frequencies and then analyze the frequency counts. The space complexity is O(n) due to the storage of frequency counts in a hashmap.

1Understanding character frequency is crucial to solving this problem.
2Recognizing the conditions under which frequencies can be equalized helps in optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.