0425MediumArrayDynamic ProgrammingMatrix

Least Strain Down the Wall

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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 training wall is bolted with holds in a square grid of n rows and n columns. holds[r][c] is the strain, in newtons, that using the hold in row r and column c costs a climber. Some holds are shaped as rests and carry a negative strain, because the climber recovers on them.

A descent uses exactly one hold from every row. It may begin at any hold in row 0 and finish at any hold in row n - 1. From a hold in column c, the only holds the climber can reach in the next row are those in columns c - 1, c and c + 1, and only where that column exists on the wall.

The cost of a descent is the total strain of the holds it uses. Return the smallest cost any descent can have.

Examples

Example 1

Input: holds = [[2, 9, 9], [9, 1, 9], [3, 9, 9]]
Output: 6

Starting at column 0 in the top row costs 2, then column 1 in the middle row costs 1, then column 0 in the bottom row costs 3, for 6 newtons in total. Each step moves at most one column sideways.

Example 2

Input: holds = [[1, 100, 100], [100, 100, 2], [100, 100, 3]]
Output: 104

Taking column 0 at the top costs 1, column 1 in the middle costs 100, and column 2 at the bottom costs 3, which is 104 newtons for a descent that shifts one column right at each step.

Example 3

Input: holds = [[-7]]
Output: -7

The wall has a single hold, and using it is the only descent there is.

Example 4

Input: holds = [[-100, -100], [-100, -100]]
Output: -200

Every hold is a rest worth -100, so any descent takes two of them and totals -200.

Example 5

Input: holds = [[0, 1, 100], [100, 0, 1], [1, 100, 0]]
Output: 0

Taking column 0 at the top, column 1 in the middle and column 2 at the bottom costs 0 + 0 + 0 = 0, and each of those steps moves one column to the right.

Constraints

n == holds.length == holds[i].length
1 <= n <= 100
-100 <= holds[i][j] <= 100

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 min_falling_path_sum(holds: list[list[int]]) -> int:

Java

public int minFallingPathSum(int[][] holds)

The intended approach

Time O(n^2) for an n by n wall: each of the n^2 holds is settled with three comparisons.Space O(n): one row of running totals plus the fresh row being filled.

Keep one row of running totals, where entry `c` is the least strain of any descent that has got as far as the current row at column `c`. Row 0 seeds it with the strains themselves. Building the next row needs only the three entries above and beside each column, so a fresh row is filled from the current one and then takes its place. The answer is the smallest entry once the bottom row has been built.

Pythonentry point: min_falling_path_sum

def min_falling_path_sum(holds):
    # Carry one row of running totals down the wall: entry c holds the least
    # strain of any descent that has reached row r at column c. Each new row is
    # built from at most three entries of the previous one and then replaces it.
    size = len(holds)
    running = list(holds[0])
    for row in range(1, size):
        fresh = [0] * size
        for col in range(size):
            least = running[col]
            if col > 0 and running[col - 1] < least:
                least = running[col - 1]
            if col + 1 < size and running[col + 1] < least:
                least = running[col + 1]
            fresh[col] = holds[row][col] + least
        running = fresh
    answer = running[0]
    for value in running:
        if value < answer:
            answer = value
    return answer

The obvious version

Walk out every descent there is from an explicit stack of partial descents, branching three ways at each row, and keep the smallest total that reaches the bottom row. Exponential in n.

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_min_falling_path_sum(holds):
    size = len(holds)
    best = None
    stack = []
    for col in range(size):
        stack.append((0, col, holds[0][col]))
    while stack:
        row, col, total = stack.pop()
        if row == size - 1:
            if best is None or total < best:
                best = total
            continue
        for step in (col - 1, col, col + 1):
            if 0 <= step < size:
                stack.append((row + 1, step, total + holds[row + 1][step]))
    return best

How this problem was checked

Reference solution executed against 14 test cases — 14 passed
5 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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