1173EasyMathEnumeration

Triples of Squares That Add Up

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Trains the technique from

LeetCode 1925Count Square Sum Triples

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.

Return how many ordered triples (a, b, c) of whole numbers between 1 and n satisfy a * a + b * b == c * c.

Triples in a different order count separately, so (3, 4, 5) and (4, 3, 5) are two of them.

Examples

Example 1

Input: n = 13
Output: 6

Three pairs of legs work within thirteen, each counted twice for its two orders: 3 with 4, 6 with 8, and 5 with 12.

Example 2

Input: n = 12
Output: 4

Two pairs work within twelve, 3 with 4 and 6 with 8, each counted twice.

Example 3

Input: n = 4
Output: 0

Nothing within four works at all.

Constraints

1 <= n <= 250

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 count_triples(n: int) -> int:

Java

public int countTriples(int n)

The intended approach

Time O(n^2) over the pairs.Space O(n) for the squares.

Once `a` and `b` are chosen, `c` is settled: it has to be the whole number whose square is `a * a + b * b`, if there is one at all and if it does not run past `n`. So the third number never needs a loop of its own. Gather the squares of `1` through `n` into a set once, then walk every pair `a` and `b` and ask whether their squares add to something in that set. That drops the work from the limit cubed to the limit squared, with a single lookup per pair. Because both orders of a pair are walked separately, each ordered triple is counted exactly once and no doubling is needed at the end.

Pythonentry point: count_triples

def count_triples(n):
    squares = set()
    for c in range(1, n + 1):
        squares.add(c * c)
    total = 0
    for a in range(1, n + 1):
        for b in range(1, n + 1):
            if a * a + b * b in squares:
                total += 1
    return total

The obvious version

Walk all three numbers over the whole range and count the triples where the first two squares add up to the third. Costs the limit cubed.

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_count_triples(n):
    total = 0
    for a in range(1, n + 1):
        for b in range(1, n + 1):
            for c in range(1, n + 1):
                if a * a + b * b == c * c:
                    total += 1
    return total

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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