How do you solve Frequency of the Most Frequent Element on LeetCode?

Frequency of the Most Frequent Element (LeetCode #1838) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting the array helps in efficiently managing the increments needed to equalize elements.

What is the time complexity of Frequency of the Most Frequent Element?

The optimal solution for Frequency of the Most Frequent Element runs in O(n log n) time and uses O(1) space. The sorting step takes O(n log n), and the sliding window approach runs in O(n), making the overall complexity dominated by the sorting step.

What is the brute force solution for Frequency of the Most Frequent Element?

The brute force approach for Frequency of the Most Frequent Element is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible value in the array and calculating how many times we can increment elements to reach that value. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Frequency of the Most Frequent Element?

Always consider edge cases, such as when k is very small or large compared to the values in the array. Explain your thought process clearly while coding; it helps the interviewer understand your approach. Practice writing clean and efficient code, as readability matters in interviews.

What are common mistakes when solving Frequency of the Most Frequent Element?

Failing to sort the array before applying the sliding window technique. Not correctly calculating the total increments needed for the current window.

#1838

Frequency of the Most Frequent Element

Medium

ArrayBinary SearchGreedySliding 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 approach uses a sliding window technique after sorting the array. By maintaining a window of elements that can be incremented to the same value, we can efficiently calculate the maximum frequency.

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Algorithm

4 steps

1Step 1: Sort the array.
2Step 2: Use two pointers to maintain a window of elements that can be made equal to the largest element in the window.
3Step 3: Calculate the total increments needed to make all elements in the window equal to the rightmost element and check if it exceeds k.
4Step 4: If it does, move the left pointer to reduce the window size. If not, update the maximum frequency.

solution.py15 lines

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

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Complexity note: The sorting step takes O(n log n), and the sliding window approach runs in O(n), making the overall complexity dominated by the sorting step.

1Sorting the array helps in efficiently managing the increments needed to equalize elements.
2Using a sliding window allows us to dynamically adjust our range of elements without recalculating from scratch.

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