Check if All Characters Have Equal Number of Occurrences Python/Java solution

In Python, a dictionary or collections.Counter is typically used to count characters and compare frequencies. In Java, a HashMap performs the same role. Both implementations iterate through the string once and then validate that all counts match.

What is the best approach for Check if All Characters Have Equal Number of Occurrences?

The best approach uses a hash table (frequency dictionary). Count occurrences of each character in one pass, then verify that all frequency values are identical. This runs in O(n) time with O(k) space where k is the number of distinct characters.

Is Check if All Characters Have Equal Number of Occurrences asked at Google/Amazon/Meta?

This type of frequency-counting string problem appears frequently in interviews at large tech companies including Amazon, Google, and Meta. It tests basic hash table usage, string traversal, and reasoning about character frequency constraints.

What data structure is used in Check if All Characters Have Equal Number of Occurrences?

The core data structure is a hash table (dictionary or map). It stores each character as a key and its frequency as the value, allowing constant-time updates and lookups during iteration.

What is the time complexity of Check if All Characters Have Equal Number of Occurrences?

The optimal solution runs in O(n) time because each character in the string is processed once while building the frequency map. Checking the frequency values takes O(k) time, where k is the number of unique characters. Overall complexity remains linear.

Description Hints 2 Solution Video 3:49

1941. Check if All Characters Have Equal Number of Occurrences

Q: Is Check if All Characters Have Equal Number of Occurrences easy or hard?

The problem is classified as Easy on LeetCode with a high acceptance rate around 79%. It mainly checks understanding of string traversal and frequency counting using a hash map.

Q: How to solve Check if All Characters Have Equal Number of Occurrences in O(n)?

Iterate through the string and store character counts in a hash map. After counting, store the first frequency and compare it against every other value in the map. If any count differs, return false; otherwise return true. Both counting and validation together run in O(n) time.

Easy79.4% acceptance4 companies

Given a string s, return true if s is a good string, or false otherwise.

A string s is good if all the characters that appear in s have the same number of occurrences (i.e., the same frequency).

Example 1:

Input: s = "abacbc"
Output: true
Explanation: The characters that appear in s are 'a', 'b', and 'c'. All characters occur 2 times in s.

Example 2:

Input: s = "aaabb"
Output: false
Explanation: The characters that appear in s are 'a' and 'b'.
'a' occurs 3 times while 'b' occurs 2 times, which is not the same number of times.

Constraints: