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

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

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

The brute force approach for Maximum Frequency of an Element After Performing Operations I is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we try every possible combination of adding values to the elements of the array within the allowed range. This helps us understand the problem but is inefficient for larger inputs. 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 I?

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

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

Always consider edge cases, such as when numOperations is 0. Explain your thought process clearly while coding. Practice similar problems to get comfortable with sliding window and sorting techniques.

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

Not sorting the array before applying the sliding window technique. Miscalculating the operations needed to adjust elements.

#3346

Maximum Frequency of an Element After Performing Operations I

Medium

ArrayBinary SearchSliding WindowSortingPrefix SumHash MapArray

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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)

By sorting the array and using a sliding window approach, we can efficiently determine the maximum frequency of any element after performing the allowed operations. This reduces the number of operations we need to check.

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Algorithm

3 steps

1Step 1: Sort the array to group similar elements together.
2Step 2: Use a sliding window to find the maximum frequency of elements that can be made equal to the rightmost element in the window.
3Step 3: Calculate the total operations needed to make all elements in the window equal to the rightmost element and check if it is within the allowed operations.

solution.py14 lines

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

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

1Sorting helps in grouping similar elements together.
2Using a sliding window allows us to efficiently calculate the operations needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.