0437MediumMathTwo PointersBinary Search

Two Square Courts From One Delivery

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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 groundworks firm has taken delivery of exactly c identical square slabs and wants to lay two square courts with them, using up every slab and cutting none.

A court is a solid square block, so a court whose side measures s slabs swallows s * s slabs. A side of zero slabs is allowed and simply means that court is not laid at all.

Return true if some choice of the two side lengths uses the delivery up exactly, and false if some slabs would always be left over.

Examples

Example 1

Input: c = 8
Output: true

Two courts of side two each swallow four slabs, which accounts for all eight.

Example 2

Input: c = 6
Output: false

Every pairing of side lengths either leaves slabs in the yard or calls for more than were delivered.

Example 3

Input: c = 26
Output: true

A court of side five swallows twenty-five slabs and a court of side one swallows the last slab.

Example 4

Input: c = 0
Output: true

Nothing was delivered, and two courts of side zero use up nothing.

Constraints

0 <= c <= 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 judge_square_sum(c: int) -> bool:

Java

public boolean judgeSquareSum(int c)

The intended approach

Time O(sqrt(c)): each step raises the smaller side or lowers the larger, and they start at most sqrt(c) apart.Space O(1): two side lengths.

Hold two candidate side lengths and squeeze them towards one another. The smaller side begins at zero and the larger at the widest court the delivery could support on its own. Compare the slabs the pair swallows with the delivery: falling short can only be fixed by widening the smaller court, and overshooting can only be fixed by narrowing the larger one, so every comparison eliminates one candidate side outright and the sweep finishes in a number of steps proportional to the square root of the delivery.

Pythonentry point: judge_square_sum

import math


def judge_square_sum(c):
    # Squeeze the two side lengths towards each other. The smaller side starts at
    # zero and the larger at the biggest side that fits the delivery on its own.
    # A shortfall can only be made up by widening the small court, and an excess
    # can only be relieved by narrowing the large one, so each comparison retires
    # one candidate side for good. O(sqrt(c)) time, O(1) space.
    small = 0
    large = math.isqrt(c)
    while small <= large:
        used = small * small + large * large
        if used == c:
            return True
        if used < c:
            small += 1
        else:
            large -= 1
    return False

The obvious version

Collect the slab counts of every square block that fits inside the delivery into a set, then for each of them look up whether the leftover slab count is also in the set.

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_judge_square_sum(c):
    blocks = set()
    side = 0
    while side * side <= c:
        blocks.add(side * side)
        side += 1
    for value in blocks:
        if c - value in blocks:
            return True
    return False

How this problem was checked

Reference solution executed against 18 test cases — 18 passed
4 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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