0263MediumArrayDynamic ProgrammingStackMonotonic Stack

Weakest Reading Per Stretch

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Trains the technique from

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 pipeline carries gauges at consecutive pumping stations. readings[i] is the pressure the gauge at station i reports.

A stretch is any block of one or more neighbouring stations. What a stretch can push through is set by its weakest gauge, so a stretch is rated at the smallest reading inside it. A pipeline of n stations therefore has n * (n + 1) / 2 stretches, each with its own rating.

Add together the ratings of all the stretches. The figure climbs fast, so give it as a remainder modulo 10^9 + 7.

Examples

Example 1

Input: readings = [4, 2, 3, 6]
Output: 28

The four single stations are rated 4, 2, 3 and 6, the three pairs are rated 2, 2 and 3, the two triples are rated 2 and 2, and the whole line is rated 2, which adds to 28.

Example 2

Input: readings = [2, 2]
Output: 6

Each station on its own is rated 2, and the stretch holding both of them is rated 2 as well, giving 6.

Example 3

Input: readings = [5]
Output: 5

A single station forms one stretch, and its rating is its own reading.

Constraints

1 <= readings.length <= 3 * 10^4
1 <= readings[i] <= 3 * 10^4
Return the total modulo 10^9 + 7

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 total_stretch_minimums(readings: list[int]) -> int:

Java

public int totalStretchMinimums(int[] readings)

The intended approach

Time O(n)Space O(n) for the stack of indices

Charge each stretch to the one station that sets its rating instead of visiting stretches one by one. Station `mid` is the charged station for every stretch that starts after the nearest earlier station whose reading is at most `readings[mid]` and ends before the nearest later station whose reading is strictly smaller, which is `(mid - left) * (right - mid)` stretches, each contributing `readings[mid]`. Taking one boundary as `at most` and the other as `strictly smaller` means a stretch whose smallest reading occurs several times is charged once and only once. A single left-to-right pass over a stack of indices with increasing readings finds both boundaries: an index leaves the stack exactly when the reading that ends its run arrives, and whatever index is left underneath it is its earlier boundary. One closing round with a sentinel below every legal reading flushes the stack.

Pythonentry point: total_stretch_minimums

def total_stretch_minimums(readings):
    limit = 10 ** 9 + 7
    n = len(readings)
    stack = []
    total = 0
    for right in range(n + 1):
        cur = 0 if right == n else readings[right]
        while stack and readings[stack[-1]] > cur:
            mid = stack.pop()
            left = stack[-1] if stack else -1
            total += readings[mid] * (mid - left) * (right - mid)
        stack.append(right)
    return total % limit

The obvious version

Fix each starting station, walk rightwards keeping the smallest reading seen so far, and add that reading once per ending station. O(n^2).

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_total_stretch_minimums(readings):
    limit = 10 ** 9 + 7
    total = 0
    n = len(readings)
    for i in range(n):
        low = readings[i]
        for j in range(i, n):
            if readings[j] < low:
                low = readings[j]
            total += low
    return total % limit

How this problem was checked

Reference solution executed against 10 test cases — 10 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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