0887MediumArrayStackMonotonic StackPrefix SumCartesian Tree

Weakest Link Times the Load

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Trains the technique from

LeetCode 1856Maximum Subarray Min-Product

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A run of couplings has strengths nums. For any contiguous stretch of it, the stretch's grade is the smallest strength in that stretch multiplied by the total of the strengths in it.

Return the largest grade over every non-empty contiguous stretch, modulo 10^9 + 7.

Examples

Example 1

Input: nums = [4, 11, 7, 2, 9, 6]
Output: 126

The stretch 11, 7 has smallest strength 7 and totals 18, giving 126. Nothing beats it: reaching one further left drops the weakest link to 4, and reaching right drops it to 2. The best stretch on the other side, 9 and 6, only reaches 90.

Example 2

Input: nums = [1, 9, 9, 9, 1]
Output: 243

The three nines in the middle have smallest strength 9 and total 27, giving 243. Reaching out to either 1 would cut the weakest link to 1.

Example 3

Input: nums = [1000000, 1000000, 1000000, 1000000]
Output: 999972007

The whole run is the best stretch, and its grade is a million times four million, which leaves this remainder once reduced.

Constraints

1 <= nums.length <= 10^5
1 <= nums[i] <= 10^6

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def max_sum_min_product(nums: list[int]) -> int:

Java

public int maxSumMinProduct(int[] nums)

The intended approach

Time O(n) for n couplings: each one is pushed and popped at most once.Space O(n) for the prefix totals and the stack.

Every stretch has some smallest strength in it, so instead of enumerating stretches, ask for each coupling what the best stretch is that has *that* coupling as its weakest link. Since all strengths are positive, growing such a stretch only raises the total, so the best one is the widest stretch on which the coupling is still the minimum: from just past the nearest strictly smaller strength on the left to just before the nearest strictly smaller on the right. Both boundaries come from one pass with a stack kept in increasing order of strength, popping whenever a smaller strength arrives, which is what makes the whole sweep linear. Prefix totals then give each window's total in constant time, and the answer is the largest product found. Compare the true products and reduce only once at the end, since reducing first would ruin the comparison. Ties need no special care: when equal strengths sit side by side, one of them gets the whole run as its window, and that is the stretch worth having.

Pythonentry point: max_sum_min_product

def max_sum_min_product(nums):
    n = len(nums)
    prefix = [0] * (n + 1)
    for i in range(n):
        prefix[i + 1] = prefix[i] + nums[i]
    left = [0] * n     # first index of the window where nums[i] is still the minimum
    stack = []
    for i in range(n):
        while stack and nums[stack[-1]] >= nums[i]:
            stack.pop()
        left[i] = stack[-1] + 1 if stack else 0
        stack.append(i)
    right = [0] * n    # last index of that window
    stack = []
    for i in range(n - 1, -1, -1):
        while stack and nums[stack[-1]] >= nums[i]:
            stack.pop()
        right[i] = stack[-1] - 1 if stack else n - 1
        stack.append(i)
    best = 0
    for i in range(n):
        total = prefix[right[i] + 1] - prefix[left[i]]
        grade = nums[i] * total
        if grade > best:
            best = grade
    return best % 1000000007

The obvious version

Walk every start, extend the stretch one coupling at a time while carrying the running total and the running smallest strength, and keep the largest product met. Quadratic in the coupling count and it never reasons about which coupling is the weakest link of which stretch.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_max_sum_min_product(nums):
    n = len(nums)
    best = 0
    for start in range(n):
        total = 0
        low = None
        for end in range(start, n):
            total += nums[end]
            if low is None or nums[end] < low:
                low = nums[end]
            grade = low * total
            if grade > best:
                best = grade
    return best % 1000000007

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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