Introduction

In this blog post, we will explore how to solve the LeetCode problem "1856" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• Approach:

  • Utilize the concept of Next Smaller Element to the Left and Right.
  • For each element in nums, find the range where it's the minimum.
  • Calculate the min-product for this range and keep track of the maximum.

• Complexity:

  • Runtime: O(n), where n is the length of the input array nums.
  • Storage: O(n)

Code

    def max_min_product(nums):
    n = len(nums)
    prefix_sum = [0] * (n + 1)
    for i in range(n):
        prefix_sum[i + 1] = prefix_sum[i] + nums[i]

    left = [0] * n
    right = [0] * n
    stack = []

    # Calculate Next Smaller Element to the Left
    for i in range(n):
        while stack and nums[stack[-1]] >= nums[i]:
            stack.pop()
        if stack:
            left[i] = stack[-1] + 1
        else:
            left[i] = 0
        stack.append(i)

    stack = []

    # Calculate Next Smaller Element to the Right
    for i in range(n - 1, -1, -1):
        while stack and nums[stack[-1]] >= nums[i]:
            stack.pop()
        if stack:
            right[i] = stack[-1] - 1
        else:
            right[i] = n - 1
        stack.append(i)

    max_product = 0
    for i in range(n):
        current_min = nums[i]
        current_sum = prefix_sum[right[i] + 1] - prefix_sum[left[i]]
        current_product = current_min * current_sum
        max_product = max(max_product, current_product)

    return max_product % (10**9 + 7)