#1838

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Frequency of the Most Frequent Element

Maximize the frequency of an element in an array by incrementing at most `k` elements. Use binary search and greedy techniques to find the solution.

array binary search greedy sliding window sorting

Problem Statement

Given an integer array nums and an integer k, you are allowed to perform at most k operations. Each operation allows you to select an index and increment the value at that index by 1. The goal is to maximize the frequency of the most frequent element in the array after performing at most k operations.

Return the maximum possible frequency of any element after performing the allowed operations. The problem can be approached by binary search over the possible frequency values, ensuring that the chosen frequency can be reached with at most k increments using a sliding window.

Examples

Example 1

Input: nums = [1,2,4], k = 5

Output: 3

Increment the first element three times and the second element two times to make nums = [4,4,4]. 4 has a frequency of 3.

Example 2

Input: nums = [1,4,8,13], k = 5

Output: 2

There are multiple optimal solutions:

Increment the first element three times to make nums = [4,4,8,13]. 4 has a frequency of 2.
Increment the second element four times to make nums = [1,8,8,13]. 8 has a frequency of 2.
Increment the third element five times to make nums = [1,4,13,13]. 13 has a frequency of 2.

Example 3

Input: nums = [3,9,6], k = 2

Output: 1

Example details omitted.

Constraints

1 <= nums.length <= 105
1 <= nums[i] <= 105
1 <= k <= 105

Solution Approach

Binary Search on Frequency

Perform a binary search over the possible frequency values to find the maximum frequency that can be achieved. Start by searching over the range of 1 to n, where n is the length of the array. For each midpoint in the binary search, check if it's possible to reach that frequency using the sliding window approach.

Sliding Window Approach

For each potential maximum frequency, use a sliding window to maintain a subarray that can be transformed into a sequence of the most frequent element with at most k operations. Keep track of the sum of differences between elements in the window and the smallest element to determine the number of operations needed.

Greedy Increment Strategy

Within the sliding window, ensure that we only increment the elements that are necessary to meet the target frequency. This greedy approach ensures that we are always making the most efficient use of the available operations to maximize the frequency of a single element.

Complexity Analysis

Metric	Value
Time	O(n \cdot \log{}n)
Space	O(n)

The time complexity of the solution is O(n * log n) due to the binary search over the frequency space and the sliding window that checks each frequency. The space complexity is O(n) because we maintain an auxiliary array for tracking the frequency of elements within the window and other intermediate results.

What Interviewers Usually Probe

Ability to combine binary search and greedy techniques to solve a problem with constraints on operations.
Demonstrates understanding of sliding window techniques in optimizing search spaces and reducing redundant computations.
Effective handling of array manipulation while maintaining clarity in approach and efficiency.

Common Pitfalls or Variants

Common pitfalls

Failing to handle the sliding window efficiently, resulting in excessive computations or incorrect frequency checks.
Misunderstanding the binary search space, leading to suboptimal choices for the maximum frequency.
Incorrectly handling the number of operations k, either exceeding it or failing to utilize it optimally.

Follow-up variants

What if the number of operations k is reduced to 0? This would change the problem to a simple frequency counting problem.
What if instead of incrementing elements, you were allowed to decrement them? This could require reversing the logic of the solution.
How would the solution change if the array contained only unique elements?

FAQ

What is the core approach for solving the Frequency of the Most Frequent Element problem?

The core approach is binary search over the frequency range combined with a sliding window technique to ensure that the number of operations used doesn't exceed k.

How does binary search help in solving this problem?

Binary search is used to explore the possible frequencies efficiently, narrowing down the search space by testing whether a given frequency can be achieved with at most k operations.

What role does the sliding window technique play in solving this problem?

The sliding window is used to find the subarray that can be transformed into the most frequent element within the allowed k operations. It efficiently checks potential solutions in linear time.

Can the problem be solved using brute force?

While a brute force approach is possible, it would be inefficient and fail to scale for large input sizes. The binary search and sliding window techniques significantly improve the solution's efficiency.

How does GhostInterview assist with this problem?

GhostInterview helps by providing a clear approach, breaking down the use of binary search and sliding window, and offering real-time problem-solving hints to avoid common pitfalls.

terminal

Solution

Solution 1: Sorting + Prefix Sum + Binary Search

According to the problem description, we can draw three conclusions:

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class Solution:
    def maxFrequency(self, nums: List[int], k: int) -> int:
        def check(m: int) -> bool:
            for i in range(m, n + 1):
                if nums[i - 1] * m - (s[i] - s[i - m]) <= k:
                    return True
            return False

        n = len(nums)
        nums.sort()
        s = list(accumulate(nums, initial=0))
        l, r = 1, n
        while l < r:
            mid = (l + r + 1) >> 1
            if check(mid):
                l = mid
            else:
                r = mid - 1
        return l

Solution 2: Sorting + Two Pointers

We can also use two pointers to maintain a sliding window, where all elements in the window can be changed to the maximum value in the window. The number of operations for the elements in the window is $s$, and $s \leq k$.

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class Solution:
    def maxFrequency(self, nums: List[int], k: int) -> int:
        def check(m: int) -> bool:
            for i in range(m, n + 1):
                if nums[i - 1] * m - (s[i] - s[i - m]) <= k:
                    return True
            return False

        n = len(nums)
        nums.sort()
        s = list(accumulate(nums, initial=0))
        l, r = 1, n
        while l < r:
            mid = (l + r + 1) >> 1
            if check(mid):
                l = mid
            else:
                r = mid - 1
        return l

Continue Practicing

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