Maximum Frequency of an Element After Performing Operations I Python/Java solution

Both Python and Java implementations typically start by sorting the array and then applying a sliding window to track the maximum group of elements that can be converted to the same value. The logic uses two pointers and a counter for how many operations are required within the window.

What is the best approach for Maximum Frequency of an Element After Performing Operations I?

The sliding window approach after sorting the array is typically the best solution. It scans the array with two pointers and maintains a window of values that can be converted to a common target within the allowed operation range. The approach runs in O(n log n) time due to sorting and uses O(1) extra space.

Is Maximum Frequency of an Element After Performing Operations I asked at Google/Amazon/Meta?

Problems involving frequency maximization, sliding window ranges, and array transformations frequently appear in interviews at companies like Amazon, Google, and Meta. Variants of this problem test sorting, two‑pointer techniques, and reasoning about limited operations.

What data structure is used in Maximum Frequency of an Element After Performing Operations I?

The solution mainly relies on arrays combined with algorithmic techniques such as sorting, sliding window, and binary search. Some implementations also use a prefix sum array to quickly compute counts within value ranges.

What is the time complexity of Maximum Frequency of an Element After Performing Operations I?

The optimal solutions run in O(n log n) time. The log n factor comes from sorting the array or performing binary searches, while the window traversal itself is linear. Space complexity ranges from O(1) for the sliding window method to O(n) when prefix sums are used.

Description Hints 1 Solution Video 37:48

3346. Maximum Frequency of an Element After Performing Operations I

Q: Is Maximum Frequency of an Element After Performing Operations I easy or hard?

The problem is rated Medium because the core logic requires recognizing that sorting enables a range-based sliding window solution. The implementation is straightforward once the window condition and operation limit are understood.

Q: How to solve Maximum Frequency of an Element After Performing Operations I in O(n)?

A pure O(n) solution is not practical because the array must usually be sorted to efficiently group values that fall within the allowed adjustment range. After sorting, a sliding window scans the array once, giving O(n log n) overall complexity. The linear pass maintains the largest valid window of convertible elements.

Medium40.1% acceptance4 companies

You are given an integer array nums and two integers k and numOperations.

You must perform an operation numOperations times on nums, where in each operation you:

Select an index i that was not selected in any previous operations.
Add an integer in the range [-k, k] to nums[i].

Return the maximum possible frequency of any element in nums after performing the operations.

Example 1:

Input: nums = [1,4,5], k = 1, numOperations = 2

Output: 2

Explanation:

We can achieve a maximum frequency of two by:

Adding 0 to nums[1]. nums becomes [1, 4, 5].
Adding -1 to nums[2]. nums becomes [1, 4, 4].

Example 2:

Input: nums = [5,11,20,20], k = 5, numOperations = 1

Output: 2

Explanation:

We can achieve a maximum frequency of two by:

Adding 0 to nums[1].

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵
0 <= k <= 10⁵
0 <= numOperations <= nums.length

Asked at 4 companies

Google Amazon Meta Bloomberg

Topics

Hints open one at a time

Hint 1

Sort the array and try each value in range as a candidate.

37:48

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