Sum of Subarray Minimums
<p>Given an array of integers arr, find the sum of <code>min(b)</code>, where <code>b</code> ranges over every (contiguous) subarray of <code>arr</code>. Since the answer may be large, return the answer <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [3,1,2,4]
<strong>Output:</strong> 17
<strong>Explanation:</strong> 
Subarrays are [3], [1], [2], [4], [3,1], [1,2], [2,4], [3,1,2], [1,2,4], [3,1,2,4]. 
Minimums are 3, 1, 2, 4, 1, 1, 2, 1, 1, 1.
Sum is 17.
</pre>

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

<pre>
<strong>Input:</strong> arr = [11,81,94,43,3]
<strong>Output:</strong> 444
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= arr[i] &lt;= 3 * 10<sup>4</sup></code></li>
</ul>
Stack; Array; Dynamic Programming; Monotonic Stack
function sumSubarrayMins(arr: number[]): number {
    const n: number = arr.length;
    const left: number[] = Array(n).fill(-1);
    const right: number[] = Array(n).fill(n);
    const stk: number[] = [];
    for (let i = 0; i < n; ++i) {
        while (stk.length > 0 && arr[stk.at(-1)] >= arr[i]) {
            stk.pop();
        }
        if (stk.length > 0) {
            left[i] = stk.at(-1);
        }
        stk.push(i);
    }

    stk.length = 0;
    for (let i = n - 1; ~i; --i) {
        while (stk.length > 0 && arr[stk.at(-1)] > arr[i]) {
            stk.pop();
        }
        if (stk.length > 0) {
            right[i] = stk.at(-1);
        }
        stk.push(i);
    }

    const mod: number = 1e9 + 7;
    let ans: number = 0;
    for (let i = 0; i < n; ++i) {
        ans += ((((i - left[i]) * (right[i] - i)) % mod) * arr[i]) % mod;
        ans %= mod;
    }
    return ans;
}
