Maximize Minimum Value After Increasing K Sublists in Python

Suppose we have a list of numbers called nums and two values, size and k. Now suppose there is an operation where we take a contiguous sublist of length size and increment every element by one. We can perform this operation k times, we have to find the largest minimum value possible in nums.

So, if the input is like nums = [2, 5, 2, 2, 7], size = 3, k = 2, then the output will be 3, as we can increase [2, 5, 2] to get [3, 6, 3, 2, 7] and then increment [6, 3, 2] to get [3, 7, 4, 3, 7], minimum is 3

To solve this, we will follow these steps −

• Define a function possible() . This will take target
• events := A list of size N, and fill with 0
• moves := 0, s := 0
• for i in range 0 to N, do
  • s := s + events[i]
  • delta := target -(A[i] + s)
  • if delta > 0, then
    • moves := moves + delta
    • s := s + delta
    • if i + size < N, then
      • events[i + size] := events[i + size] - delta
• return true when moves <= K
• From the main method, do the following
• N := size of A
• left := 0, right := 10^10
• while left < right, do
  • mid :=(left + right + 1) / 2
  • if possible(mid), then
    • left := mid
  • otherwise,
    • right := mid - 1
• return left

Let us see the following implementation to get better understanding −

Example

 Live Demo

class Solution:
   def solve(self, A, size, K):
      N = len(A)
   def possible(target):
      events = [0] * N
      moves = s = 0
      for i in range(N):
         s += events[i]
         delta = target - (A[i] + s)
         if delta > 0:
            moves += delta
            s += delta
            if i + size < N:
               events[i + size] -= delta
               return moves <= K
               left, right = 0, 10 ** 10
               while left < right:
                  mid = (left + right + 1)//2
               if possible(mid):
                  left = mid
               else:
                  right = mid - 1
      return left
ob = Solution()
nums = [2, 5, 2, 2, 7]
size = 3
k = 2
print(ob.solve(nums, size, k))

Input

[2, 5, 2, 2, 7], 3, 2

Output

3