1793. Maximum Score of a Good Subarray

Tags:

• Hard
• two pointer
• greedy

Link

class Solution:
    def maximumScore(self, nums: List[int], k: int) -> int:

        res = nums[k]

        l, r = k , k

        currentMin = nums[k]

        while l != 0 or r != len(nums) - 1:

            left = 0 if l - 1 == -1  else nums[l-1]

            right = 0 if r + 1 == len(nums) else nums[r+1]

            if left > right:
                currentMin = min(currentMin, left)
                l -= 1
                res = max(res, currentMin * (r - l + 1))
            else:
                currentMin = min(currentMin, right)
                r += 1
                res = max(res, currentMin * (r - l + 1))

        return res