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

The optimal solution for Maximum Frequency of an Element After Performing Operations II runs in O(n log n) time and uses O(1) space. The sorting step takes O(n log n) time, and the sliding window traversal takes O(n) time, making this approach efficient overall.

What is the brute force solution for Maximum Frequency of an Element After Performing Operations II?

The brute force approach for Maximum Frequency of an Element After Performing Operations II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying every possible combination of operations on the array to determine the maximum frequency of any element. This is straightforward but inefficient as it checks all possibilities. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Frequency of an Element After Performing Operations II?

Maximum Frequency of an Element After Performing Operations II uses the following concepts: Array, Binary Search, Sliding Window, Sorting, Prefix Sum. The recommended patterns to study are: Sliding Window, Sorting.

What are the interview tips for Maximum Frequency of an Element After Performing Operations II?

Always think about edge cases, such as when numOperations is 0 or when k is 0. Explain your thought process clearly as you work through the problem, especially when transitioning from brute force to optimal solutions. Practice similar problems to get comfortable with patterns like sliding window and sorting.

What are common mistakes when solving Maximum Frequency of an Element After Performing Operations II?

Not considering the need to sort the array before applying the sliding window technique. Failing to correctly calculate the operations needed to make elements equal.

#3347

Maximum Frequency of an Element After Performing Operations II

Hard

ArrayBinary SearchSliding WindowSortingPrefix SumSliding WindowSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

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Intuition

Time O(n log n)Space O(1)

The optimal solution uses sorting and a sliding window approach to efficiently determine the maximum frequency of any element. By sorting the array, we can focus on making elements equal to a target value using the least operations.

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Algorithm

3 steps

1Step 1: Sort the array nums.
2Step 2: Use a sliding window to find the maximum frequency of elements that can be made equal to nums[right] using the allowed operations.
3Step 3: For each right index, calculate the total operations needed to make all elements in the current window equal to nums[right]. If it exceeds numOperations, move the left index to reduce the window size.

solution.py17 lines

1# Full working Python code
2
3def maxFrequency(nums, k, numOperations):
4    nums.sort()
5    left = 0
6    total = 0
7    max_freq = 0
8    for right in range(len(nums)):
9        total += nums[right]
10        while nums[right] * (right - left + 1) - total > numOperations:
11            total -= nums[left]
12            left += 1
13        max_freq = max(max_freq, right - left + 1)
14    return max_freq
15
16# Example usage
17print(maxFrequency([1, 4, 5], 1, 2))

ℹ

Complexity note: The sorting step takes O(n log n) time, and the sliding window traversal takes O(n) time, making this approach efficient overall.

1Sorting the array allows us to efficiently determine how many elements can be made equal to a target value.
2Using a sliding window helps us keep track of the number of operations needed dynamically.

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