Maximize Element Frequency

Given an array of positive integers nums and a positive integer k, you may perform at most k operations.
In one operation, choose one index and increase nums[i] by 1.
Return the maximum possible frequency of any value in the array after performing the operations. The frequency of a value is the number of times it appears.

Function
maxFrequency(nums: int[], k: int) → int

Examples
Example 1
nums = [1,3,5,7,8,9,10,15]
k = 6
return = 4
Increase 7 three times, 8 twice, and 9 once. The array then contains four occurrences of 10, using exactly 6 operations.
Example 2
nums = [1,2,4]
k = 5
return = 3
Increase 1 to 4 using 3 operations and increase 2 to 4 using 2 operations. All three elements become 4.
Example 3
nums = [1,4,8,13]
k = 5
return = 2
Two elements can be made equal, for example by increasing 8 to 13. Making any three elements equal requires more than 5 operations.
Example 4
nums = [3,9,6]
k = 2
return = 1
No value can reach another array value with only 2 increments, so the maximum frequency remains 1.

Constraints
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^5
1 <= k <= 10^5

Sort the array. Increments only go up, so the best target for any group is the largest element in that group, and the group should be a contiguous run in sorted order. Slide a window with right pointer r. The cost to lift everything in the window to nums[r] is nums[r] * windowSize - windowSum. While that cost exceeds k, move the left pointer and shrink the sum. Track the max window size. That's O(n log n) for the sort and O(n) for the scan. The common pitfall is the hash map approach, which counts existing frequencies and ignores that the target may already be in the array but groups still need contiguous neighbors. Another trap is integer overflow in nums[r] * size, so use 64-bit in languages that need it. If the window condition slips from your head mid-OA, StealthCoder can hand you the template while the proctor sees nothing.

If you see this problem in your OA tomorrow, the play is to recognize the pattern in 30 seconds. StealthCoder buys you that recognition.