<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>

# 84. Largest Rectangle in Histogram
# 2334. Subarray With Elements Greater Than Varying Threshold
class Solution:
    def maxSumMinProduct(self, nums: List[int]) -> int:
        nums += [0]
        n = len(nums)
        mod = int(10**9 + 7)

        # for every num in nums, find its next smaller element on both left and right
        # then we know the longest subarray that num is min value

        prefix_sum = list(itertools.accumulate(nums)) + [0]
        ans = 0
        stk = [-1]

        for i, num in enumerate(nums):
            while stk and nums[stk[-1]] > num:
                subarray_min_val = nums[stk.pop()]
                # i is the right bound of subarray
                # so prefix sum should be from i-1 to stk[-1]
                # edge case handled: when stk[-1] == -1, prefix_sum[-1] = 0

                subarray_sum = prefix_sum[i - 1] - prefix_sum[stk[-1]]

                ans = max(ans, subarray_min_val * subarray_sum)
            # end while
            stk.append(i)
        # end for

        return ans % mod

29000