0295MediumMathDynamic ProgrammingBreadth-First SearchKnapsack ProblemComplete Knapsack

Fewest Square Tiles

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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 courtyard crew paves a strip whose surface measures exactly area square metres. The supplier stocks square slabs in every whole-metre side length: 1 by 1, 2 by 2, 3 by 3 and so on, in unlimited quantity, and a slab of side s covers s * s square metres.

The crew lays slabs so that the covered surface adds up to area with nothing left over and nothing overlapping. Only the areas matter here, not how the slabs sit next to each other, and the same side length may be used as often as needed.

Given area, return the smallest number of slabs whose areas add up to exactly area.

Examples

Example 1

Input: area = 28
Output: 4

Three slabs of side 2 and one of side 4 cover 4 + 4 + 4 + 16, which is 28 square metres with four slabs.

Example 2

Input: area = 18
Output: 2

Two slabs of side 3 cover 9 + 9, which is exactly 18 square metres.

Constraints

1 <= area <= 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 fewest_square_tiles(area: int) -> int:

Java

public int fewestSquareTiles(int area)

The intended approach

Time O(area * sqrt(area))Space O(area)

Fill a table from 1 up to the target where entry a holds the fewest slabs covering exactly a square metres. Each entry starts at a, the count for all unit slabs, and is then relaxed against one plus the entry for a minus every square that fits. The final entry is the answer.

Pythonentry point: fewest_square_tiles

def fewest_square_tiles(area):
    # dp[a] is the fewest tiles that cover exactly a square metres.  Every a is
    # relaxed through each tile area that fits inside it, so the work is the sum of
    # sqrt(a) over all a.
    # O(area * sqrt(area)) time, O(area) space.
    dp = [0] * (area + 1)
    for a in range(1, area + 1):
        best = a                    # a tiles of side 1 always cover a
        side = 1
        while side * side <= a:
            cand = dp[a - side * side] + 1
            if cand < best:
                best = cand
            side += 1
        dp[a] = best
    return dp[area]

The obvious version

Breadth-first search over the surface still uncovered, subtracting each usable square area at every step, and report the depth at which zero is first reached.

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 collections import deque


def bf_fewest_square_tiles(area):
    squares = []
    side = 1
    while side * side <= area:
        squares.append(side * side)
        side += 1
    seen = {area}
    queue = deque([(area, 0)])
    while queue:
        rem, used = queue.popleft()
        if rem == 0:
            return used
        for q in squares:
            if q > rem:
                break
            nxt = rem - q
            if nxt not in seen:
                seen.add(nxt)
                queue.append((nxt, used + 1))
    return -1

How this problem was checked

Reference solution executed against 12 test cases — 12 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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