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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

• Java
• C++
• Python
• Go
• TypeScript

• 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
  }
  

• 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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