README_EN.md

1856. Maximum Subarray Min-Product

中文文档

Description

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 109 + 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 <= 105
• 1 <= nums[i] <= 107

Solutions

Python3

class Solution:
    def maxSumMinProduct(self, nums: List[int]) -> int:
        n = len(nums)
        pre_sum = [0] * (n + 1)
        for i in range(1, n + 1):
            pre_sum[i] = pre_sum[i - 1] + nums[i - 1]

        stack = []
        next_lesser = [n] * n
        for i in range(n):
            while stack and nums[stack[-1]] > nums[i]:
                next_lesser[stack.pop()] = i
            stack.append(i)

        stack = []
        prev_lesser = [-1] * n
        for i in range(n - 1, -1, -1):
            while stack and nums[stack[-1]] > nums[i]:
                prev_lesser[stack.pop()] = i
            stack.append(i)

        res = 0
        for i in range(n):
            start, end = prev_lesser[i], next_lesser[i]
            t = nums[i] * (pre_sum[end] - pre_sum[start + 1])
            res = max(res, t)
        return res % (10 ** 9 + 7)

Java

class Solution {
    public int maxSumMinProduct(int[] nums) {
        int n = nums.length;
        long[] preSum = new long[n + 1];
        for (int i = 1; i < n + 1; ++i) {
            preSum[i] = preSum[i - 1] + nums[i - 1];
        }
        Deque<Integer> stack = new ArrayDeque<>();
        int[] nextLesser = new int[n];
        Arrays.fill(nextLesser, n);
        for (int i = 0; i < n; ++i) {
            while (!stack.isEmpty() && nums[stack.peek()] > nums[i]) {
                nextLesser[stack.pop()] = i;
            }
            stack.push(i);
        }

        stack = new ArrayDeque<>();
        int[] prevLesser = new int[n];
        Arrays.fill(prevLesser, -1);
        for (int i = n - 1; i >= 0; --i) {
            while (!stack.isEmpty() && nums[stack.peek()] > nums[i]) {
                prevLesser[stack.pop()] = i;
            }
            stack.push(i);
        }
        long res = 0;
        for (int i = 0; i < n; ++i) {
            int start = prevLesser[i], end = nextLesser[i];
            long t = nums[i] * (preSum[end] - preSum[start + 1]);
            res = Math.max(res, t);
        }
        return (int) (res % 1000000007);
    }
}

...