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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.

  • 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;
    }
}

  • // 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

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