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]. nums becomes [1, 4, 5].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:
nums[1].
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 1050 <= k <= 1050 <= numOperations <= nums.lengthThe sliding window technique can be utilized here to keep track of a window of elements that can be turned into the same value by using the allowed operations efficiently.
Sort the array first, then utilize a variable to track the cost of making elements within the current window equal. Expand the window while the cost is within the allowed operations, and shrink the window otherwise.
Here, we use a sliding window strategy. By sorting the input array, we calculate a cost to bring each window of elements to the same value by adding or subtracting using available operations. We adjust the window size effectively to stay within the given operations limit to maximize the frequency.
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Time Complexity: O(n log n), due to the sorting operation. Space Complexity: O(1), as no extra space is used besides input sorting itself.
Another approach to solve this problem is using Binary Search combined with computing prefix sums. We attempt to make each element the highest frequency possible by using binary search on frequency and verify using prefix sum to ensure that allowed operations suffice to make all elements in range equal.
By prefix sum and binary search, checking the frequency transformable within the limit of allowed operations is done. This involves iterating through and checking transformations using the prefix sum accumulated to aid efficient calculation of needed operations per hypothetical frequency.
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Time Complexity: O(n log m), where m is the max element - due to binary search. Space Complexity: O(n) for storing prefix sum.
| Approach | Complexity |
|---|---|
| Approach 1: Sliding Window Technique | Time Complexity: O(n log n), due to the sorting operation. Space Complexity: O(1), as no extra space is used besides input sorting itself. |
| Approach 2: Binary Search with Prefix Sum | Time Complexity: O(n log m), where m is the max element - due to binary search. Space Complexity: O(n) for storing prefix sum. |
Frequency of the Most Frequent Element - Sliding Window - Leetcode 1838 • NeetCode • 97,819 views views
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