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 is2) has a min-product of2 * (3+2+5) = 2 * 10 = 20.
Given an array of integers nums, return the maximum min-product of any non-empty subarray ofnums. Since the answer may be large, return it modulo109 + 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.
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.
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.
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.
1 <= nums.length <= 1051 <= nums[i] <= 107
implSolution{pubfnmax_sum_min_product(nums:Vec<i32>) -> i32{let n = nums.len();letmut prefix_sum = vec![0; n + 1];letmut stack = vec![];letmut indexl = vec![None; n];letmut indexr = vec![None; n];letmut ret = nums[0]asi64* nums[0]asi64;for i in0..n { prefix_sum[i + 1] = prefix_sum[i] + nums[i]asi64;whileletSome(&j) = stack.last(){if nums[j] >= nums[i]{ stack.pop();}else{ indexl[i] = Some(j);break;}} stack.push(i);} stack.clear();for i in(0..n).rev(){whileletSome(&j) = stack.last(){if nums[j] >= nums[i]{ stack.pop();}else{ indexr[i] = Some(j);break;}} stack.push(i);letmut sum = prefix_sum[indexr[i].unwrap_or(n)];ifletSome(j) = indexl[i]{ sum -= prefix_sum[j + 1];} ret = ret.max(nums[i]asi64* sum);}(ret % 1_000_000_007)asi32}}