<p>You are given a <strong>0-indexed</strong> array <code>maxHeights</code> of <code>n</code> integers.</p>

<p>You are tasked with building <code>n</code> towers in the coordinate line. The <code>i<sup>th</sup></code> tower is built at coordinate <code>i</code> and has a height of <code>heights[i]</code>.</p>

<p>A configuration of towers is <strong>beautiful</strong> if the following conditions hold:</p>

<ol>
	<li><code>1 &lt;= heights[i] &lt;= maxHeights[i]</code></li>
	<li><code>heights</code> is a <strong>mountain</strong> array.</li>
</ol>

<p>Array <code>heights</code> is a <strong>mountain</strong> if there exists an index <code>i</code> such that:</p>

<ul>
	<li>For all <code>0 &lt; j &lt;= i</code>, <code>heights[j - 1] &lt;= heights[j]</code></li>
	<li>For all <code>i &lt;= k &lt; n - 1</code>, <code>heights[k + 1] &lt;= heights[k]</code></li>
</ul>

<p>Return <em>the <strong>maximum possible sum of heights</strong> of a beautiful configuration of towers</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> maxHeights = [5,3,4,1,1]
<strong>Output:</strong> 13
<strong>Explanation:</strong> One beautiful configuration with a maximum sum is heights = [5,3,3,1,1]. This configuration is beautiful since:
- 1 &lt;= heights[i] &lt;= maxHeights[i]  
- heights is a mountain of peak i = 0.
It can be shown that there exists no other beautiful configuration with a sum of heights greater than 13.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> maxHeights = [6,5,3,9,2,7]
<strong>Output:</strong> 22
<strong>Explanation:</strong> One beautiful configuration with a maximum sum is heights = [3,3,3,9,2,2]. This configuration is beautiful since:
- 1 &lt;= heights[i] &lt;= maxHeights[i]
- heights is a mountain of peak i = 3.
It can be shown that there exists no other beautiful configuration with a sum of heights greater than 22.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> maxHeights = [3,2,5,5,2,3]
<strong>Output:</strong> 18
<strong>Explanation:</strong> One beautiful configuration with a maximum sum is heights = [2,2,5,5,2,2]. This configuration is beautiful since:
- 1 &lt;= heights[i] &lt;= maxHeights[i]
- heights is a mountain of peak i = 2. 
Note that, for this configuration, i = 3 can also be considered a peak.
It can be shown that there exists no other beautiful configuration with a sum of heights greater than 18.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n == maxHeights.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= maxHeights[i] &lt;= 10<sup>9</sup></code></li>
</ul>

r'''
# Requirements

n = len(maxHeights)
1 <= n <= 100,000
1 <= maxHeights[i] <= 1,000,000,000

# Solutions

1. Brute force
    - O(n^2) time, O(1) auxiliary space
2. Monotonic stack
    - O(n) time, O(n) auxiliary space
'''
class Solution:
    def maximumSumOfHeights(self, max_heights: list[int]) -> int:
        n = len(max_heights)
        left_sums, right_sums = [0] * n, [0] * n

        # Find sum of heights to the left
        stack = []

        for j in range(n):
            curr_height = max_heights[j]

            while stack and curr_height <= stack[-1][0]:
                stack.pop()

            left_pos = stack[-1][1] if stack else -1
            left_sums[j] = (left_sums[left_pos] if left_pos >= 0 else 0) + curr_height * (j - left_pos)

            if not stack or curr_height > stack[-1][0]:
                stack.append((curr_height, j))

        # Find sum of height to the right, and the max sum of heights
        stack, max_sum = [], 0

        for k in range(n - 1, -1, -1):
            curr_height = max_heights[k]

            while stack and curr_height <= stack[-1][0]:
                stack.pop()

            right_pos = stack[-1][1] if stack else n
            right_sums[k] = (right_sums[right_pos] if right_pos < n else 0) + curr_height * (right_pos - k)
            max_sum = max(max_sum, left_sums[k] + right_sums[k] - curr_height)

            if not stack or curr_height > stack[-1][0]:
                stack.append((curr_height, k))

        return max_sum

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