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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

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

  • class Solution {
public:
    int countTriples(int n) {
        int res = 0;
        for (int a = 1; a <= n; ++a) {
            for (int b = 1; b <= n; ++b) {
                int t = a * a + b * b;
                int c = (int) sqrt(t);
                if (c <= n && c * c == t) {
                    ++res;
                }
            }
        }
        return res;
    }
};

  • class Solution:
    def countTriples(self, n: int) -> int:
        res = 0
        for a in range(1, n + 1):
            for b in range(1, n + 1):
                t = a**2 + b**2
                c = int(sqrt(t))
                if c <= n and c**2 == t:
                    res += 1
        return res


  • func countTriples(n int) int {
	res := 0
	for a := 1; a <= n; a++ {
		for b := 1; b <= n; b++ {
			t := a*a + b*b
			c := int(math.Sqrt(float64(t)))
			if c <= n && c*c == t {
				res++
			}
		}
	}
	return res
}

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