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1925. Count Square Sum Triples

Description

A square triple (a,b,c) is a triple where a, b, and c are integers and a2 + b2 = c2.

Given an integer n, return the number of square triples such that 1 <= a, b, c <= n.

 

Example 1:

Input: n = 5
Output: 2
Explanation: The square triples are (3,4,5) and (4,3,5).

Example 2:

Input: n = 10
Output: 4
Explanation: The square triples are (3,4,5), (4,3,5), (6,8,10), and (8,6,10).

 

Constraints:

  • 1 <= n <= 250

Solutions

Solution 1: Enumeration

We enumerate $a$ and $b$ in the range $[1, n)$, then calculate $c = \sqrt{a^2 + b^2}$. If $c$ is an integer and $c \leq n$, then we have found a Pythagorean triplet, and we increment the answer by one.

After the enumeration is complete, return the answer.

The time complexity is $O(n^2)$, where $n$ is the given integer. The space complexity is $O(1)$.

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class Solution:
    def countTriples(self, n: int) -> int:
        ans = 0
        for a in range(1, n):
            for b in range(1, n):
                x = a * a + b * b
                c = int(sqrt(x))
                if c <= n and c * c == x:
                    ans += 1
        return ans

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class Solution {
    public int countTriples(int n) {
        int ans = 0;
        for (int a = 1; a < n; a++) {
            for (int b = 1; b < n; b++) {
                int x = a * a + b * b;
                int c = (int) Math.sqrt(x);
                if (c <= n && c * c == x) {
                    ans++;
                }
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int countTriples(int n) {
        int ans = 0;
        for (int a = 1; a < n; ++a) {
            for (int b = 1; b < n; ++b) {
                int x = a * a + b * b;
                int c = static_cast<int>(sqrt(x));
                if (c <= n && c * c == x) {
                    ++ans;
                }
            }
        }
        return ans;
    }
};

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func countTriples(n int) (ans int) {
    for a := 1; a < n; a++ {
        for b := 1; b < n; b++ {
            x := a*a + b*b
            c := int(math.Sqrt(float64(x)))
            if c <= n && c*c == x {
                ans++
            }
        }
    }
    return
}

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function countTriples(n: number): number {
    let ans = 0;
    for (let a = 1; a < n; a++) {
        for (let b = 1; b < n; b++) {
            const x = a * a + b * b;
            const c = Math.floor(Math.sqrt(x));
            if (c <= n && c * c === x) {
                ans++;
            }
        }
    }
    return ans;
}

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