0300EasyMathBinary Search

Square Crate Block

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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 depot has crates identical crates and wants to set them out as one solid square block on the floor: the same number of rows as columns, every position filled, and not a single crate left over or borrowed.

Given crates, return true when a whole number of rows exists that makes such a block, and false when it does not.

Settle it with arithmetic on integers. Do not reach for a square-root or power routine from the standard library, and do not float the count through a fractional type on the way.

Examples

Example 1

Input: crates = 808201
Output: true

899 rows of 899 crates each use exactly 808201 crates.

Example 2

Input: crates = 11
Output: false

A 3 by 3 block uses 9 and leaves 2 spare, and a 4 by 4 block would need 5 crates more, so no square block fits 11 exactly.

Constraints

1 <= crates <= 2^31 - 1

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 fits_square_block(crates: int) -> bool:

Java

public boolean fitsSquareBlock(int crates)

The intended approach

Time O(log n)Space O(1)

The side length lies between 1 and the crate count itself, and squaring is increasing over that range, so a halving search settles it. Square the midpoint each round, answer true on an exact hit, and move whichever end of the range the comparison rules out.

Pythonentry point: fits_square_block

def fits_square_block(crates):
    # Search the side length between 1 and the whole count, halving the range on
    # each comparison of the squared candidate against the crate count.
    # O(log n) time, O(1) space.
    lo = 1
    hi = crates
    while lo <= hi:
        mid = (lo + hi) // 2
        area = mid * mid
        if area == crates:
            return True
        if area < crates:
            lo = mid + 1
        else:
            hi = mid - 1
    return False

The obvious version

Peel off successive odd numbers 1, 3, 5, ... from the count, since those partial sums are exactly the squares, and check whether the count lands on zero.

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_fits_square_block(crates):
    left = crates
    step = 1
    while left > 0:
        left -= step
        step += 2
    return left == 0

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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