What is the brute force solution for Check if All Characters Have Equal Number of Occurrences?

The brute force approach for Check if All Characters Have Equal Number of Occurrences is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks the frequency of each character by comparing each character's count with every other character's count. This is straightforward but inefficient as it involves multiple comparisons. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check if All Characters Have Equal Number of Occurrences?

Check if All Characters Have Equal Number of Occurrences uses the following concepts: Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Counting.

What are the interview tips for Check if All Characters Have Equal Number of Occurrences?

Always think about the simplest approach first, then optimize. Explain your thought process clearly as you code; it shows your understanding. Consider edge cases and test your solution with different inputs.

What are common mistakes when solving Check if All Characters Have Equal Number of Occurrences?

Not considering edge cases, such as strings with only one character. Using inefficient methods to count occurrences, leading to unnecessary time complexity.

#1941

Check if All Characters Have Equal Number of Occurrences

Easy

Hash TableStringCountingHash MapCounting

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 solution uses a hash map to count the occurrences of each character in a single pass, and then checks if all counts are the same. This is efficient and reduces the number of operations significantly.

⚙️

Algorithm

3 steps

1Step 1: Create a hash map to store the frequency of each character.
2Step 2: Iterate through the string and populate the hash map with character counts.
3Step 3: Extract the frequency values from the hash map and check if they are all the same.

solution.py5 lines

1from collections import Counter
2
3def areOccurrencesEqual(s):
4    freq = Counter(s)
5    return len(set(freq.values())) == 1

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once to count characters and then check the counts, which is efficient. The space complexity is O(n) due to the storage of character frequencies.

1Using a hash map allows for efficient counting of character occurrences in a single pass.
2Checking if all values in a hash map are the same can be done using a set, simplifying the logic.

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