Test/add leetcode 1941 tests

problems/easy/CheckIfAllCharactersHaveEqualNumberOfOccurrences1941/solution.py

@@ -0,0 +1,23 @@
+# Tags: String, Hash Table
+from collections import Counter
+
+
+class Solution:
+    def are_occurrences_equal(self, s: str) -> bool:
+        """
+        Time complexity: O(n)
+        - n represents the length of the input string s
+        - Counter(s) traverses the entire string once: O(n)
+        - count_table.values() extracts frequency values: O(k), k = unique chars
+        - set() removes duplicates: O(k)
+        - len() calculates set size: O(1)
+        - Since k <= n, overall: O(n)
+
+        Space complexity: O(k)
+        - k represents the number of unique characters
+        - Counter stores k character-frequency pairs: O(k)
+        - The set of unique frequencies: O(k)
+        - Overall: O(k), where k <= n
+        """
+        count_table = Counter(s)
+        return len(set(count_table.values())) == 1

problems/easy/CheckIfAllCharactersHaveEqualNumberOfOccurrences1941/test_solution.py

@@ -0,0 +1,14 @@
+import pytest
+from tests.base_test import BaseTestSolution
+from .solution import Solution
+
+
+class TestSolution(BaseTestSolution):
+    solution = Solution()
+
+    @pytest.mark.parametrize("method_name, s, expected, timeout", [
+        ('are_occurrences_equal', "abacbc", True, None),
+        ('are_occurrences_equal', "aaabb", False, None),
+    ])
+    def test_are_occurrences_equal(self, method_name, s, expected, timeout):
+        self._run_test(self.solution, method_name, (s,), expected, timeout)