0287MediumArrayDynamic Programming

Sensor Cascade Minimum

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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 drift monitor prints its readings as a growing stack of rows. Row 0 holds one reading, and every later row holds exactly one more reading than the row above it, so row i holds i + 1 readings. Each reading is a signed drift in millidegrees and may be negative.

A trace begins on the single reading in row 0 and ends on some reading in the last row. From reading j of row i, a trace may continue to reading j or to reading j + 1 of row i + 1. The weight of a trace is the sum of every reading it lands on, including the first and the last.

Given rows, return the smallest weight any trace can have.

Examples

Example 1

Input: rows = [[3],[1,7],[10,10,0]]
Output: 10

The trace 3 -> 7 -> 0 starts on the only reading of row 0, moves to reading 1 of row 1 and then to reading 2 of row 2, so every hop is allowed. Its readings add to 10.

Example 2

Input: rows = [[2],[5,-3],[1,-1,4],[0,2,-6,3]]
Output: -8

The trace 2 -> -3 -> -1 -> -6 keeps to indices 0, 1, 1, 2, and each hop either holds the index or raises it by one. The four readings add to -8.

Constraints

1 <= rows.length <= 200
rows[0].length == 1
rows[i].length == rows[i - 1].length + 1
-10^4 <= rows[i][j] <= 10^4

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 cheapest_cascade(rows: list[list[int]]) -> int:

Java

public int cheapestCascade(List<List<Integer>> rows)

The intended approach

Time O(m^2)Space O(m)

Work from the bottom row upward. Keep one scratch row whose entry j is the smallest total obtainable starting at reading j of the row currently being folded, and rewrite it in place for the row above by adding that reading to the smaller of the two readings it can reach. The answer is the single surviving entry for row 0.

Pythonentry point: cheapest_cascade

def cheapest_cascade(rows):
    # Fold the stack upward.  best[j] holds the smallest achievable total from
    # reading j of the row currently being processed down to the bottom row.  A
    # single scratch row is reused; entry best[j + 1] is still the value from the
    # row below when best[j] is overwritten, because j increases.
    # O(m^2) time over the m*(m+1)/2 readings, O(m) space.
    best = list(rows[-1])
    for i in range(len(rows) - 2, -1, -1):
        row = rows[i]
        for j in range(len(row)):
            down = best[j] if best[j] < best[j + 1] else best[j + 1]
            best[j] = row[j] + down
    return best[0]

The obvious version

Enumerate every hold-or-advance decision sequence for the m - 1 hops, add up the readings each one lands on, and keep the smallest total.

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

from itertools import product


def bf_cheapest_cascade(rows):
    m = len(rows)
    best = None
    for steps in product((0, 1), repeat=m - 1):
        j = 0
        total = rows[0][0]
        for i, s in enumerate(steps, start=1):
            j += s
            total += rows[i][j]
        if best is None or total < best:
            best = total
    return best

How this problem was checked

Reference solution executed against 8 test cases — 8 passed
2 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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