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New Problem Solution - "1856. Maximum Subarray Min-Product"
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README.md

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| # | Title | Solution | Difficulty |
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|---| ----- | -------- | ---------- |
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|1856|[Maximum Subarray Min-Product](https://leetcode.com/problems/maximum-subarray-min-product/) | [C++](./algorithms/cpp/maximumSubarrayMinProduct/MaximumSubarrayMinProduct.cpp)|Medium|
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|1855|[Maximum Distance Between a Pair of Values](https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/) | [C++](./algorithms/cpp/maximumDistanceBetweenAPairOfValues/MaximumDistanceBetweenAPairOfValues.cpp)|Medium|
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|1854|[Maximum Population Year](https://leetcode.com/problems/maximum-population-year/) | [C++](./algorithms/cpp/maximumPopulationYear/MaximumPopulationYear.cpp)|Easy|
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|1851|[Minimum Interval to Include Each Query](https://leetcode.com/problems/minimum-interval-to-include-each-query/) | [C++](./algorithms/cpp/minimumIntervalToIncludeEachQuery/MinimumIntervalToIncludeEachQuery.cpp)|Hard|
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// Source : https://leetcode.com/problems/maximum-subarray-min-product/
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// Author : Hao Chen
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// Date : 2021-05-09
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/*****************************************************************************************************
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*
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* The min-product of an array is equal to the minimum value in the array multiplied by the array's
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* sum.
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*
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* For example, the array [3,2,5] (minimum value is 2) has a min-product of 2 * (3+2+5) = 2 *
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* 10 = 20.
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*
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* Given an array of integers nums, return the maximum min-product of any non-empty subarray of nums.
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* Since the answer may be large, return it modulo 10^9 + 7.
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*
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* Note that the min-product should be maximized before performing the modulo operation. Testcases are
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* generated such that the maximum min-product without modulo will fit in a 64-bit signed integer.
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*
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* A subarray is a contiguous part of an array.
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*
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* Example 1:
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*
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* Input: nums = [1,2,3,2]
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* Output: 14
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* Explanation: The maximum min-product is achieved with the subarray [2,3,2] (minimum value is 2).
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* 2 * (2+3+2) = 2 * 7 = 14.
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*
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* Example 2:
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*
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* Input: nums = [2,3,3,1,2]
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* Output: 18
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* Explanation: The maximum min-product is achieved with the subarray [3,3] (minimum value is 3).
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* 3 * (3+3) = 3 * 6 = 18.
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*
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* Example 3:
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*
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* Input: nums = [3,1,5,6,4,2]
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* Output: 60
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* Explanation: The maximum min-product is achieved with the subarray [5,6,4] (minimum value is 4).
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* 4 * (5+6+4) = 4 * 15 = 60.
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*
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* Constraints:
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*
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* 1 <= nums.length <= 10^5
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* 1 <= nums[i] <= 10^7
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******************************************************************************************************/
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class Solution {
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public:
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int maxSumMinProduct(vector<int>& nums) {
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nums.push_back(0); //edge case
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//prefix sum
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vector<long> sums(nums.size(), 0);
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// sums[i] = sum (num[0], num[1], num[2], ... num[n-1])
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// sums[m] - sums[n] = sum (num[n], sum[n+1] .... nums[m-1]); m > n
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for(int i=0; i<nums.size()-1; i++) {
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sums[i+1] = sums[i] + nums[i];
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}
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stack<int> s;
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long m = 0;
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for(int i=0; i<nums.size(); i++) {
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while( !s.empty() && nums[s.top()] > nums[i]) {
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int min = nums[s.top()]; s.pop();
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int start = s.empty() ? 0 : s.top() + 1;
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int end = i;
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m = max(m , min * (sums[end] - sums[start]));
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// cout << "[";
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// for(int k = start; k < end-1; k++) {
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// cout << nums[k] << ",";
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// }
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// cout << nums[end-1] << "], " << min << "*" << (sums[end] - sums[start])
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// << "=" << min * (sums[end] - sums[start]) << endl;
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}
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// if the num is increasing, then push into stack
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s.push(i);
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}
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//cout << endl;
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return m % 1000000007;
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}
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};

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