feat (day 10): add squares_with_three counter with optimal generator solution

Author: JohnSunny21

FreeCodeCamp/CodingQ/threeStrikes.py

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+"""
+3 Strikes
+Given an integer between 1 and 10,000, return a count of how many numbers from 1 up to that integer whose square contains at least one digit 3.
+
+
+O/P : ==>
+
+squares_with_three(1) should return 0.
+"""
+
+def squares_with_three(n):
+    count = 0
+
+    for i in range(1, n+1):
+        num = i ** 2
+        if '3' in str(num):
+            count += 1
+    return count
+
+
+# Optimal Optimization
+# If the code ever need to be optimized and slightly faster for large n,
+# we could avoid recomputing str(num) twice:
+
+def squares_with_three_optimal(n):
+    return sum('3' in str(i * i) for i in range(1,n+1))
+# This uses a generator expression and is just as readable.
+

FreeCodeCamp/CodingQ/threeStrikesTest.py

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+import unittest
+from threeStrikes import squares_with_three
+
+class threeStrikesTest(unittest.TestCase):
+
+    def test1(self):
+        self.assertEqual(squares_with_three(1),0)
+
+    def test2(self):
+        self.assertEqual(squares_with_three(10),1)
+
+    def test3(self):
+        self.assertEqual(squares_with_three(100),19)
+
+    def test4(self):
+        self.assertEqual(squares_with_three(1000),326)
+
+    def test5(self):
+        self.assertEqual(squares_with_three(10000),4531)
+
+
+if __name__ == "__main__":
+    unittest.main()