Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/1925.html

1925. Count Square Sum Triples

Level

Easy

Description

A square triple (a,b,c) is a triple where a, b, and c are integers and a^2 + b^2 = c^2.

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

Solution

Loop over a, b and c and count the number of square triples.

• Java
• C++
• Python
• Go

• class Solution {
      public int countTriples(int n) {
          int count = 0;
          for (int a = 1; a <= n; a++) {
              for (int b = 1; b <= n; b++) {
                  for (int c = Math.max(a, b) + 1; c <= n; c++) {
                      if (a * a + b * b == c * c)
                          count++;
                  }
              }
          }
          return count;
      }
  }
  
  ############
  
  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;
      }
  }
  

• // OJ: https://leetcode.com/problems/count-square-sum-triples/
  // Time: O(N^2)
  // Space: O(1)
  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 cs = a * a + b * b, c = sqrt(cs);
                  if (c > n) break;
                  if (c * c == cs) ++ans;
              }
          }
          return ans;
      }
  };
  

• 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
  
  ############
  
  # 1925. Count Square Sum Triples
  # https://leetcode.com/problems/count-square-sum-triples/
  
  class Solution:
      def countTriples(self, n: int) -> int:
          res = 0
          mp = collections.defaultdict(int)
  
          for a in range(1, n + 1):
              for b in range(1, n + 1):
                  mp[a * a + b * b] += 1
          
          for c in range(1, n + 1):
              res += mp[c * c]
  
          
          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
  }
  

All Problems

All Solutions