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

README.md

@@ -9,6 +9,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|1856|[Maximum Subarray Min-Product](https://leetcode.com/problems/maximum-subarray-min-product/) | [C++](./algorithms/cpp/maximumSubarrayMinProduct/MaximumSubarrayMinProduct.cpp)|Medium|
 |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|
 |1854|[Maximum Population Year](https://leetcode.com/problems/maximum-population-year/) | [C++](./algorithms/cpp/maximumPopulationYear/MaximumPopulationYear.cpp)|Easy|
 |1851|[Minimum Interval to Include Each Query](https://leetcode.com/problems/minimum-interval-to-include-each-query/) | [C++](./algorithms/cpp/minimumIntervalToIncludeEachQuery/MinimumIntervalToIncludeEachQuery.cpp)|Hard|

algorithms/cpp/maximumSubarrayMinProduct/MaximumSubarrayMinProduct.cpp

@@ -0,0 +1,85 @@
+// Source : https://leetcode.com/problems/maximum-subarray-min-product/
+// Author : Hao Chen
+// Date   : 2021-05-09
+
+/***************************************************************************************************** 
+ *
+ * The min-product of an array is equal to the minimum value in the array multiplied by the array's 
+ * sum.
+ * 
+ * 	For example, the array [3,2,5] (minimum value is 2) has a min-product of 2 * (3+2+5) = 2 * 
+ * 10 = 20.
+ * 
+ * Given an array of integers nums, return the maximum min-product of any non-empty subarray of nums. 
+ * Since the answer may be large, return it modulo 10^9 + 7.
+ * 
+ * Note that the min-product should be maximized before performing the modulo operation. Testcases are 
+ * generated such that the maximum min-product without modulo will fit in a 64-bit signed integer.
+ * 
+ * A subarray is a contiguous part of an array.
+ * 
+ * Example 1:
+ * 
+ * Input: nums = [1,2,3,2]
+ * Output: 14
+ * Explanation: The maximum min-product is achieved with the subarray [2,3,2] (minimum value is 2).
+ * 2 * (2+3+2) = 2 * 7 = 14.
+ * 
+ * Example 2:
+ * 
+ * Input: nums = [2,3,3,1,2]
+ * Output: 18
+ * Explanation: The maximum min-product is achieved with the subarray [3,3] (minimum value is 3).
+ * 3 * (3+3) = 3 * 6 = 18.
+ * 
+ * Example 3:
+ * 
+ * Input: nums = [3,1,5,6,4,2]
+ * Output: 60
+ * Explanation: The maximum min-product is achieved with the subarray [5,6,4] (minimum value is 4).
+ * 4 * (5+6+4) = 4 * 15 = 60.
+ * 
+ * Constraints:
+ * 
+ * 	1 <= nums.length <= 10^5
+ * 	1 <= nums[i] <= 10^7
+ ******************************************************************************************************/
+
+class Solution {
+public:
+    int maxSumMinProduct(vector<int>& nums) {
+        nums.push_back(0); //edge case
+
+        //prefix sum
+        vector<long> sums(nums.size(), 0);
+        // sums[i] = sum (num[0], num[1], num[2], ... num[n-1])
+        // sums[m] - sums[n] = sum (num[n], sum[n+1] .... nums[m-1]);  m > n
+        for(int i=0; i<nums.size()-1; i++) {
+            sums[i+1] = sums[i] + nums[i];
+        }
+
+        stack<int> s;
+        long m = 0;
+        for(int i=0; i<nums.size(); i++) {
+            while( !s.empty() && nums[s.top()] > nums[i]) {
+                int min = nums[s.top()]; s.pop();
+                int start = s.empty() ? 0 : s.top() + 1;
+                int end = i;
+                m = max(m , min * (sums[end] - sums[start]));
+
+                // cout << "[";
+                // for(int k = start; k < end-1; k++) {
+                //    cout << nums[k] << ",";
+                // }
+                // cout << nums[end-1] << "], " << min << "*" << (sums[end] - sums[start]) 
+                //     << "=" << min * (sums[end] - sums[start]) << endl;
+
+            }
+            // if the num is increasing, then push into stack
+            s.push(i);
+        }
+        //cout << endl;
+        return m % 1000000007;
+
+    }
+};