Maximum Frequency of an Element After Performing Operations I

Problem Description

Ah, the classic dilemma of wanting to be the most popular kid in school! Imagine you have a bag of candies (or any other item you fancy), and you want to make sure that one particular type of candy is the most frequent one in your collection. But wait! You can only change the type of candy a limited number of times. Sounds like a fun party game, right?

In the Maximum Frequency of an Element After Performing Operations I problem, you are given an array of integers (nums), and you can perform a limited number of operations (numOperations) to increase the frequency of any number in the array. The catch? You can only increase a number by k at a time. So, how do you maximize the frequency of your favorite candy?

Solution in Python


from sortedcontainers import SortedDict
import collections

class Solution:
    def maxFrequency(self, nums: list[int], k: int, numOperations: int) -> int:
        ans = 1
        adjustable = 0
        count = collections.Counter(nums)
        line = SortedDict()
        candidates = set()

        for num in nums:
            line[num - k] = line.get(num - k, 0) + 1
            line[num + k + 1] = line.get(num + k + 1, 0) - 1
            candidates.add(num)
            candidates.add(num - k)
            candidates.add(num + k + 1)

        for num in sorted(candidates):
            adjustable += line.get(num, 0)
            adjusted = adjustable - count[num]
            ans = max(ans, count[num] + min(numOperations, adjusted))

        return ans

Approach Explanation

The code uses a combination of a SortedDict and a Counter to keep track of the frequency of each number in the array. The idea is to iterate through potential candidates for the maximum frequency and calculate how many operations are needed to make that number the most frequent.

Counting Frequencies: It first counts how many times each number appears in the array.
Adjustable Range: It then creates a range of numbers that can be adjusted by k and keeps track of how many adjustments can be made.
Max Frequency Calculation: Finally, it calculates the maximum frequency possible by considering the number of operations available.

Time and Space Complexity

Complexity Type	Complexity
Time Complexity	O(n log n)
Space Complexity	O(n)

Real-World Example

Imagine you are at a party with a variety of snacks, and you want to ensure that your favorite snack (let’s say, nachos) is the most abundant. You can trade some of the other snacks for nachos, but you can only make a limited number of trades. The problem is essentially about maximizing the number of nachos you have after making the allowed trades.