[LeetCode Sync] Runtime - 87 ms (86.69%), Memory - 17.7 MB (54.08%)

Author: NinePiece2

Solutions/2037-count-square-sum-triples/README.md

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+<p>A <strong>square triple</strong> <code>(a,b,c)</code> is a triple where <code>a</code>, <code>b</code>, and <code>c</code> are <strong>integers</strong> and <code>a<sup>2</sup> + b<sup>2</sup> = c<sup>2</sup></code>.</p>
+
+<p>Given an integer <code>n</code>, return <em>the number of <strong>square triples</strong> such that </em><code>1 &lt;= a, b, c &lt;= n</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 5
+<strong>Output:</strong> 2
+<strong>Explanation</strong>: The square triples are (3,4,5) and (4,3,5).
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 10
+<strong>Output:</strong> 4
+<strong>Explanation</strong>: The square triples are (3,4,5), (4,3,5), (6,8,10), and (8,6,10).
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 250</code></li>
+</ul>

Solutions/2037-count-square-sum-triples/solution.py

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+class Solution:
+    def countTriples(self, n: int) -> int:
+        # result = 0
+        # for a in range(1, n):
+        #     for b in range(1, n):
+        #         val = a**2 + b**2
+        #         c = int(sqrt(val))
+        #         if c <= n and val == c**2:
+        #             result += 1
+        # return result
+
+        result = 0
+        squares = set([i**2 for i in range(1, n + 1)])
+        for a in range(1, n + 1):
+            for b in range(a + 1, n + 1):
+                if a**2 + b**2 in squares:
+                    result += 2
+        return result