In context to LeetCode 1838, the constraint says that it is only possible to increment an element in an effort to creating an element with most frequency.
The frequency of an element is the number of times it occurs in an array.
You are given an integer array nums and an integer k. In one operation, you can choose an index of nums and increment the element at that index by 1.
Return the maximum possible frequency of an element after performing at most k operations.
This is the most basic solution to it:
def maxFrequency(self, nums: List[int], k: int) -> int:
nums.sort()
left=0
max_freq,currwindowsum=0,0
for right in range(len(nums)):
currwindowsum+=nums[right]
while nums[right]*(right-left+1) > currwindowsum+k:
currwindowsum-=nums[left]
left+=1
max_freq=max(max_freq, right-left+1)
return max_freq
How to implement it, if decrements were possible??
example for the case nums=[1,2,4] k=3, The above code gives output as 2. But with decrements possible, the actual answer must be 3, where the element 1 can be formed 3 times. How to implement it.
3
When both increments and decrements are allowed, the best value in a subarray to move all elements toward is the median (rather than the largest element). The sliding window solution shown can be modified to maintain the sum of elements below the median and the sum above the median while the rest of the logic remains mostly the same.
Sample implementation:
def maxFrequency(nums: list[int], k: int) -> int:
nums.sort()
above = below = left = max_freq = 0
for right, value in enumerate(nums):
above += value
if right - left & 1:
above -= nums[left + right >> 1]
else:
below += nums[left + right >> 1]
while above - below > k:
below -= nums[left]
left += 1
if right - left & 1:
above -= nums[left + right >> 1]
else:
below += nums[left + right >> 1]
max_freq = max(max_freq, right - left + 1)
return max_freq
0