描述
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.
Note:
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^7
解析
根据题意，数组的最小乘积等于数组中的最小值乘以数组的总和。
- 例如，数组 [3,2,5]（最小值为 2）的最小乘积为 2 * (3+2+5) = 2 * 10 = 20。
给定一个整数数组 nums，返回 nums 的任何非空子数组的最大的最小乘积。 由于答案可能很大，将其取模 10^9 + 7 返回。需要注意的是，在执行模运算之前应最大化最小乘积。
最暴力的解法就是 O(n^2) ，两层循环找出所有的子数组，然后找出每个子数组的最小值与和的乘积，最后找出最大值即可，但是这方法肯定会超时，因为题目条件说明了数组的长度可能为 10^5 。因为每个元素都是大于 0 的正整数，所以我们可以找出以每个元素 nums[i] 为最小值的最大子数组的左右边际，这个过程可以通过单调栈的思想来解决，然后使用前缀和求出这个左右边际范围内的子数组的和与 nums[i] 相乘，最后经过比较找出最大值，对 (10**9+7) 取模即可返回答案。
解答
class Solution(object):
def maxSumMinProduct(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
N = len(nums)
nextSmaller = [N] * N
preSmaller = [-1] * N
stack = []
for i in range(N):
while stack and nums[stack[-1]] > nums[i]:
nextSmaller[stack.pop()] = i
stack.append(i)
stack = []
for i in range(N-1,-1,-1):
while stack and nums[stack[-1]] > nums[i]:
preSmaller[stack.pop()] = i
stack.append(i)
presum = [0] * N
presum[0] = nums[0]
for i in range(1, N):
presum[i] = presum[i-1] + nums[i]
result = 0
for i in range(N):
a = 0 if preSmaller[i] == -1 else presum[preSmaller[i]]
b = presum[nextSmaller[i]-1]
result = max(result, (b-a) * nums[i])
return result%(10**9+7)
运行结果
Runtime: 1416 ms, faster than 41.67% of Python online submissions for Maximum Subarray Min-Product.
Memory Usage: 28.9 MB, less than 41.67% of Python online submissions for Maximum Subarray Min-Product.
原题链接：leetcode.com/problems/ma…
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