Is Maximum Equal Frequency easy or hard?

Maximum Equal Frequency is classified as a hard problem because it requires recognizing specific frequency patterns that allow equal counts after removing one element. The implementation is short but the reasoning behind the valid cases is non-trivial.

Maximum Equal Frequency Python/Java solution

Python and Java implementations typically use a dictionary or HashMap for counting numbers and another map for frequency-of-frequency tracking. The algorithm iterates once through the array and checks valid frequency configurations at each prefix.

How to solve Maximum Equal Frequency in O(n)?

Iterate through the array while maintaining two maps: one for element frequency and another for how many elements have a given frequency. Track the current maximum frequency and check whether the prefix can form equal frequencies after removing one element. Each prefix check is constant time, leading to O(n) total complexity.

What is the best approach for Maximum Equal Frequency?

The best approach uses two hash maps to track element frequencies and how many numbers share the same frequency. While scanning the array once, you update these counts and check a few valid patterns that allow equal frequencies after removing one element. This solution runs in O(n) time and O(n) space.

Is Maximum Equal Frequency asked at Google/Amazon/Meta?

Maximum Equal Frequency is a hard-level frequency counting problem similar to questions asked at companies like Google, Amazon, and Meta. It tests hash map usage, reasoning about frequency distributions, and careful handling of edge cases.

What data structure is used in Maximum Equal Frequency?

Hash maps are the primary data structure. One map stores the frequency of each number, and another tracks how many numbers share the same frequency. This structure allows constant-time updates and quick validation of prefix conditions.

What is the time complexity of Maximum Equal Frequency?

The optimal solution runs in O(n) time because the array is processed once and each step performs constant-time hash map updates. Space complexity is O(n) in the worst case to store frequency counts of distinct numbers.

Description Hints 2 Solution Video 7:54

1224. Maximum Equal Frequency

Hard38.8% acceptance1 company

Given an array nums of positive integers, return the longest possible length of an array prefix of nums, such that it is possible to remove exactly one element from this prefix so that every number that has appeared in it will have the same number of occurrences.

If after removing one element there are no remaining elements, it's still considered that every appeared number has the same number of ocurrences (0).

Example 1:

Input: nums = [2,2,1,1,5,3,3,5]
Output: 7
Explanation: For the subarray [2,2,1,1,5,3,3] of length 7, if we remove nums[4] = 5, we will get [2,2,1,1,3,3], so that each number will appear exactly twice.

Example 2:

Input: nums = [1,1,1,2,2,2,3,3,3,4,4,4,5]
Output: 13

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

Asked at 1 company

American Express

Topics

Array Hash Table

Hints open one at a time

Hint 1

Keep track of the min and max frequencies.

Hint 2

The number to be eliminated must have a frequency of 1, same as the others or the same +1.

7:54

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