Watch 10 video solutions for Maximum Frequency of an Element After Performing Operations II, a hard level problem involving Array, Binary Search, Sliding Window. This walkthrough by Aryan Mittal has 11,002 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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:
i that was not selected in any previous operations.[-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:
nums[1], after which nums becomes [1, 4, 5].nums[2], after which 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:
nums[1].
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 1090 <= k <= 1090 <= numOperations <= nums.lengthProblem Overview: You are given an array where operations can adjust values within a certain range. The goal is to maximize the frequency of a single element after performing at most k operations. Each operation allows modifying values so that multiple elements can be aligned to the same target value.
The challenge is determining which value should become the target and how many surrounding elements can be transformed into it while staying within the allowed operation budget. Efficient solutions rely on sorting, range counting, and window-based cost tracking.
Approach 1: Sliding Window Technique (Sorting + Prefix Cost Tracking) (Time: O(n log n), Space: O(1))
Sort the array so values are processed in increasing order. Once sorted, treat each value as a potential target and expand a sliding window that represents elements you want to convert to this target. As the right pointer grows, compute the cost required to raise all elements in the window to match the current value. If the cost exceeds the allowed operations, shrink the window from the left.
The key insight: after sorting, the cheapest way to increase frequency is to raise smaller elements toward a larger target. Maintaining a running window lets you check feasibility in constant time per step. This approach combines sorting with a classic sliding window pattern to efficiently track the maximum achievable frequency.
Approach 2: Greedy with Frequency Count (Binary Search + Prefix Sum) (Time: O(n log n), Space: O(n))
This method focuses on evaluating each unique value as a potential target. After sorting the array, build prefix sums so the cost of converting a range can be computed instantly. For each index, use binary search to determine the farthest left boundary where the cost to convert all numbers to the current value does not exceed the allowed operations.
The prefix sums allow quick computation of the total increments required to raise a subarray to the target value. Greedy reasoning ensures that expanding toward smaller values gives the largest possible group. This approach blends prefix sum range calculations with binary search to efficiently test valid windows.
Recommended for interviews: The sliding window solution is typically expected. It demonstrates strong understanding of sorted array properties, two‑pointer window expansion, and incremental cost maintenance. A brute force approach would check every possible target and range, which quickly becomes quadratic. Showing that baseline reasoning helps, but implementing the optimized sliding window proves algorithmic maturity.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Sliding Window Technique (Sorted Array) | O(n log n) | O(1) | Best general solution after sorting. Efficient for large arrays and common interview expectation. |
| Greedy with Prefix Sum + Binary Search | O(n log n) | O(n) | Useful when prefix sums are already available or when analyzing valid ranges via binary search. |
| Naive Range Checking (Conceptual Baseline) | O(n^2) | O(1) | Only helpful for understanding the brute force idea before optimizing. |