Python程序:最大化提升K个子列表后的最小值
假设我们有一个名为nums的数字列表和两个值size和k。现在假设有一个操作,我们取一个长度为size的连续子列表并将每个元素加一。我们可以执行此操作k次,我们必须找到nums中可能的最大最小值。
因此,如果输入类似于nums = [2, 5, 2, 2, 7],size = 3,k = 2,则输出将为3,因为我们可以将[2, 5, 2]增加到[3, 6, 3, 2, 7],然后将[6, 3, 2]增加到[3, 7, 4, 3, 7],最小值为3
为了解决这个问题,我们将遵循以下步骤:
- 定义一个函数possible()。这将接收目标值target。
- events := 一个大小为N的列表,并填充0
- moves := 0, s := 0
- 对于范围0到N的i,执行:
- s := s + events[i]
- delta := target -(A[i] + s)
- 如果delta > 0,则
- moves := moves + delta
- s := s + delta
- 如果i + size < N,则
- events[i + size] := events[i + size] - delta
- 当moves <= K时返回true
- 从主方法中执行以下操作:
- N := A的大小
- left := 0, right := 10^10
- 当left < right时,执行:
- mid :=(left + right + 1) / 2
- 如果possible(mid)为真,则
- left := mid
- 否则,
- right := mid - 1
- 返回left
让我们看看下面的实现以更好地理解:
示例
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))
输入
[2, 5, 2, 2, 7], 3, 2
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输出
3
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