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Problem

You are given an array of integers nums (0-indexed) and an integer k.

The score of a subarray (i, j) is defined as min(nums[i], nums[i+1], ..., nums[j]) * (j - i + 1). A good subarray is a subarray where i <= k <= j.

Return the maximum possible score of a good subarray.

 

Example 1:

Input: nums = [1,4,3,7,4,5], k = 3Output: 15Explanation: The optimal subarray is (1, 5) with a score of min(4,3,7,4,5) * (5-1+1) = 3 * 5 = 15.

Example 2:

Input: nums = [5,5,4,5,4,1,1,1], k = 0Output: 20Explanation: The optimal subarray is (0, 4) with a score of min(5,5,4,5,4) * (4-0+1) = 4 * 5 = 20.

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 2 * 104
  • 0 <= k < nums.length

Sol

在區間對的時候才更新答案

class Solution:
    def maximumScore(self, nums: List[int], k: int) -> int:
        stk = []
        ret = 0
        for i,n in enumerate(nums+[0]):
            while stk and nums[stk[-1]] >= n:
                j = stk.pop()
                if stk and stk[-1]<k<i:
                    ret = max(ret, nums[j]*(i-stk[-1]-1))
                elif not stk and k<i:
                    ret = max(ret, nums[j]*i)
            stk += i,
        return ret