1801-1803 Liech buckle novice！

Just write a question every day，然后写下能让自己看懂的answer吧!
The new year has passed5All over，Write this at one time5God。

first of all1801.1802.1803，The daily questions of these three days are actually a high -quality weekly match last year，Even the simple label is also ordinary difficulty。

If the method is solved, it is a stack of a triangle，Stop increased after reaching the boundary（If it continues to increase, it will lead to timeout） 。

When both sides are reached，Just lay all the remaining bricks directly on the top layer。

1802. Specify the maximum value of the lower bid in the bounded array
      medium
      182
      company
      Amazon
      company
      Facebook
      Give you three positive integers n、index and maxSum 。You need to construct an array that meets all the conditions at the same time nums（Bidding from 0 start count）：

nums.length == n
nums[i] yes Positive integer ，in 0 <= i < n
abs(nums[i] - nums[i+1]) <= 1 ，in 0 <= i < n-1
nums 中所有元素之and不超过 maxSum
nums[index] Value maximize
Return to the array you constructed nums[index] 。

Notice：abs(x) equal x 的前提yes x >= 0 ；otherwise，abs(x) equal -x 。



Exemplary example 1：

enter：n = 4, index = 2,  maxSum = 6
Output：2
explain：Array [1,1,2,1] and [1,2,2,1] Meet all conditions。不存在其他在指定BiddingPlace具有更大值的有效Array。
Exemplary example 2：

enter：n = 6, index = 1,  maxSum = 10
Output：3


hint：

1 <= n <= maxSum <= 109
0 <= index < n

[]

class Solution:
    def maxValue(self, n: int, index: int, maxSum: int) -> int:
        diff = maxSum - n
        left = index
        right = index
        res = 1
        dl = 0
        dr = 0
        while diff > 0:  # When there are remaining bricks
            left -= 1
            right += 1
            if left >= 0:
                dl = dl + 1  # Before reaching the left boundary
            if right < n:
                dr = dr + 1  # Have not arrived at the right boundary
            if left < 0 and right >= n:  # 当到达左边界and右边界时 Exit
                # res+=diff%n==0?diff/n:diff/n+1 #Divide the remaining bricks，Exit directly
                res += diff // n if diff % n == 0 else diff // n + 1
                return res
            res += 1  # Layer renewal
            diff -= (dl + dr + 1)  # The number of bricks needed to strictly pile the top layer into strict triangle(On the left+Need to be on the right+indexPlace1indivual)

        return res