Maximum Subarray Min-Product
<p>The <strong>min-product</strong> of an array is equal to the <strong>minimum value</strong> in the array <strong>multiplied by</strong> the array&#39;s <strong>sum</strong>.</p>

<ul>
	<li>For example, the array <code>[3,2,5]</code> (minimum value is <code>2</code>) has a min-product of <code>2 * (3+2+5) = 2 * 10 = 20</code>.</li>
</ul>

<p>Given an array of integers <code>nums</code>, return <em>the <strong>maximum min-product</strong> of any <strong>non-empty subarray</strong> of </em><code>nums</code>. Since the answer may be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>Note that the min-product should be maximized <strong>before</strong> performing the modulo operation. Testcases are generated such that the maximum min-product <strong>without</strong> modulo will fit in a <strong>64-bit signed integer</strong>.</p>

<p>A <strong>subarray</strong> is a <strong>contiguous</strong> part of an array.</p>

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

<pre>
<strong>Input:</strong> nums = [1,<u>2,3,2</u>]
<strong>Output:</strong> 14
<strong>Explanation:</strong> The maximum min-product is achieved with the subarray [2,3,2] (minimum value is 2).
2 * (2+3+2) = 2 * 7 = 14.
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,<u>3,3</u>,1,2]
<strong>Output:</strong> 18
<strong>Explanation:</strong> The maximum min-product is achieved with the subarray [3,3] (minimum value is 3).
3 * (3+3) = 3 * 6 = 18.
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,1,<u>5,6,4</u>,2]
<strong>Output:</strong> 60
<strong>Explanation:</strong> The maximum min-product is achieved with the subarray [5,6,4] (minimum value is 4).
4 * (5+6+4) = 4 * 15 = 60.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>7</sup></code></li>
</ul>
Stack; Array; Prefix Sum; Monotonic Stack
class Solution {
public:
    int maxSumMinProduct(vector<int>& nums) {
        int n = nums.size();
        vector<int> left(n, -1);
        vector<int> right(n, n);
        stack<int> stk;
        for (int i = 0; i < n; ++i) {
            while (!stk.empty() && nums[stk.top()] >= nums[i]) {
                stk.pop();
            }
            if (!stk.empty()) {
                left[i] = stk.top();
            }
            stk.push(i);
        }
        stk = stack<int>();
        for (int i = n - 1; ~i; --i) {
            while (!stk.empty() && nums[stk.top()] > nums[i]) {
                stk.pop();
            }
            if (!stk.empty()) {
                right[i] = stk.top();
            }
            stk.push(i);
        }
        long long s[n + 1];
        s[0] = 0;
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        long long ans = 0;
        for (int i = 0; i < n; ++i) {
            ans = max(ans, nums[i] * (s[right[i]] - s[left[i] + 1]));
        }
        const int mod = 1e9 + 7;
        return ans % mod;
    }
};