feat(day 121): implement most frequent element finder using Counter for efficient frequency analysis

Author: JohnSunny21

FreeCodeCamp/CodingQ/MostFrequent.py

@@ -0,0 +1,42 @@
+"""
+Most Frequent
+Given an array of elements, return the element that appears most frequently.
+
+There will always be a single most frequent element.
+"""
+
+import unittest
+
+class MostFrequentTest(unittest.TestCase):
+
+    def test1(self):
+        self.assertEqual(most_frequent(["a","b","a","c"]), "a")
+
+    def test2(self):
+        self.assertEqual(most_frequent([2, 3, 5, 2, 6, 3, 2, 7, 2, 9]), 2)
+
+    def test3(self):
+        self.assertEqual(most_frequent([True, False, "False", "True", False]), False)
+        
+    def test4(self):
+        self.assertEqual(most_frequent([40, 20, 70, 30, 10, 40, 10, 50, 40, 60]), 40)
+
+from collections import Counter
+def most_frequent(arr):
+
+    freq = Counter(arr)
+    # Here we dont' need the lambda explicitly - Counter has a .get() , so you can write:
+    # return max(freq, key=freq.get)
+    return max(freq, key=lambda i : freq[i])
+    """
+    Guarantee: Since the problem states there will always be a single most frequent element, we don't need to handle ties.
+    If ties were possible, we'd need extra logic.
+    """
+
+
+
+
+if __name__ == "__main__":
+    print(most_frequent(["S","u","n","n","y"]))
+    unittest.main()
+