Program to maximize the 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