Beautiful Towers II
<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>
Stack; Array; Monotonic Stack
function maximumSumOfHeights(maxHeights: number[]): number {
    const n = maxHeights.length;
    const stk: number[] = [];
    const left: number[] = Array(n).fill(-1);
    const right: number[] = Array(n).fill(n);
    for (let i = 0; i < n; ++i) {
        const x = maxHeights[i];
        while (stk.length && maxHeights[stk.at(-1)] > x) {
            stk.pop();
        }
        if (stk.length) {
            left[i] = stk.at(-1);
        }
        stk.push(i);
    }
    stk.length = 0;
    for (let i = n - 1; ~i; --i) {
        const x = maxHeights[i];
        while (stk.length && maxHeights[stk.at(-1)] >= x) {
            stk.pop();
        }
        if (stk.length) {
            right[i] = stk.at(-1);
        }
        stk.push(i);
    }
    const f: number[] = Array(n).fill(0);
    const g: number[] = Array(n).fill(0);
    for (let i = 0; i < n; ++i) {
        const x = maxHeights[i];
        if (i && x >= maxHeights[i - 1]) {
            f[i] = f[i - 1] + x;
        } else {
            const j = left[i];
            f[i] = x * (i - j) + (j >= 0 ? f[j] : 0);
        }
    }
    for (let i = n - 1; ~i; --i) {
        const x = maxHeights[i];
        if (i + 1 < n && x >= maxHeights[i + 1]) {
            g[i] = g[i + 1] + x;
        } else {
            const j = right[i];
            g[i] = x * (j - i) + (j < n ? g[j] : 0);
        }
    }
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        ans = Math.max(ans, f[i] + g[i] - maxHeights[i]);
    }
    return ans;
}
