1100MediumArrayDynamic Programming

Finishing a Row of Panels

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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 row of panels must each be given one of three finishes. prices[i][j] is what it costs to put finish j on panel i. Two panels standing next to each other may not share a finish.

Return the least it can cost to finish the whole row.

Examples

Example 1

Input: prices = [[15, 3, 15], [14, 14, 4], [12, 2, 17]]
Output: 9

Give the first panel the second finish for 3, the middle panel the third finish for 4 and the last panel the second finish for 2. No two neighbours match and the bill comes to 9.

Example 2

Input: prices = [[8, 6, 2]]
Output: 2

One panel has no neighbour, so the cheapest of its three finishes wins.

Example 3

Input: prices = [[1, 2, 3], [1, 20, 20]]
Output: 3

Taking the cheapest finish on the first panel would push the second panel onto a finish costing 20. Paying 2 on the first panel lets the second cost 1.

Constraints

1 <= prices.length <= 100
prices[i].length == 3
1 <= prices[i][j] <= 20

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_cost(prices: list[list[int]]) -> int:

Java

public int minCost(int[][] prices)

The intended approach

Time O(n) over the panels.Space O(1) for the three running totals.

Walking the row and always grabbing the cheapest finish falls over, because a cheap choice now can force an expensive one next door. What settles the row is that only the finish on the panel just done matters to the panel in hand, and nothing further back does. So carry three running totals down the row: the least it can cost to finish everything up to the current panel, once for each finish that panel might be wearing. Moving to the next panel, each of its three totals is that panel's own price plus the cheaper of the two running totals whose finish differs from it. That is the whole step, and it takes a fixed amount of work per panel. Start the three totals at the first panel's own three prices, since it has no neighbour behind it, and the answer is the smallest of the three totals once the row runs out.

Pythonentry point: min_cost

def min_cost(prices):
    a = prices[0][0]
    b = prices[0][1]
    c = prices[0][2]
    for i in range(1, len(prices)):
        na = prices[i][0] + min(b, c)
        nb = prices[i][1] + min(a, c)
        nc = prices[i][2] + min(a, b)
        a = na
        b = nb
        c = nc
    return min(a, b, c)

The obvious version

Spell out every way of handing finishes to the panels, throw away the ones where two neighbours match, add up what is left and keep the cheapest. Costs three to the power of the panel count.

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_cost(prices):
    from itertools import product
    n = len(prices)
    best = None
    for combo in product(range(3), repeat=n):
        ok = True
        for i in range(1, n):
            if combo[i] == combo[i - 1]:
                ok = False
                break
        if not ok:
            continue
        total = 0
        for i in range(n):
            total += prices[i][combo[i]]
        if best is None or total < best:
            best = total
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 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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