refactor: migrate LeetCode 1207 to new directory structure with tests and docs

README.md

@@ -41,11 +41,11 @@ Each problem follows this structure:
 
 ## Solutions
 
-📊 **Statistics**: 90 problems solved (migrated to new structure)
+📊 **Statistics**: 91 problems solved (migrated to new structure)
 
 | Difficulty | Count |
 |------------|-------|
-| Easy       | 62    |
+| Easy       | 63    |
 | Medium     | 26    |
 | Hard       | 2     |
 

docs/EASY.md

@@ -1,6 +1,6 @@
 # Easy Problems
 
-Total: 62 problems solved
+Total: 63 problems solved
 
 ## Solutions
 
@@ -39,6 +39,7 @@ Total: 62 problems solved
 | 977 | [Squares of a Sorted Array](https://leetcode.com/problems/squares-of-a-sorted-array/) | [solution.py](../problems/easy/squares_of_a_sorted_array_977/solution.py) | [test_solution.py](../problems/easy/squares_of_a_sorted_array_977/test_solution.py) |
 | 1119 | [Remove Vowels from a String](https://leetcode.com/problems/remove-vowels-from-a-string/) | [solution.py](../problems/easy/remove_vowels_from_a_string_1119/solution.py) | [test_solution.py](../problems/easy/remove_vowels_from_a_string_1119/test_solution.py) |
 | 1160 | [Find Words That Can Be Formed by Characters](https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/) | [solution.py](../problems/easy/find_words_that_can_be_formed_by_characters_1160/solution.py) | [test_solution.py](../problems/easy/find_words_that_can_be_formed_by_characters_1160/test_solution.py) |
+| 1207 | [Unique Number of Occurrences](https://leetcode.com/problems/unique-number-of-occurrences/) | [solution.py](../problems/easy/unique_number_of_occurrences_1207/solution.py) | [test_solution.py](../problems/easy/unique_number_of_occurrences_1207/test_solution.py) |
 | 1323 | [Maximum 69 Number](https://leetcode.com/problems/maximum-69-number/) | [solution.py](../problems/easy/maximum_69_number_1323/solution.py) | [test_solution.py](../problems/easy/maximum_69_number_1323/test_solution.py) |
 | 1365 | [How Many Numbers Are Smaller Than the Current Number](https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/) | [solution.py](../problems/easy/how_many_numbers_are_smaller_than_the_current_number_1365/solution.py) | [test_solution.py](../problems/easy/how_many_numbers_are_smaller_than_the_current_number_1365/test_solution.py) |
 | 1389 | [Create Target Array in the Given Order](https://leetcode.com/problems/create-target-array-in-the-given-order/) | [solution.py](../problems/easy/create_target_array_in_the_given_order_1389/solution.py) | [test_solution.py](../problems/easy/create_target_array_in_the_given_order_1389/test_solution.py) |

problems/easy/unique_number_of_occurrences_1207/solution.py

@@ -0,0 +1,40 @@
+# Tags: Array, Hash Table
+
+from collections import Counter
+
+
+class Solution:
+    def unique_occurrences(self, arr: list[int]) -> bool:
+        """
+        Check whether every distinct value in arr has a unique number of occurrences.
+
+        Args:
+            arr: Input list of integers.
+
+        Returns:
+            True if no two distinct values occur the same number of times,
+            otherwise False.
+
+        Time complexity: O(n)
+        - n is the length of arr, k is the number of distinct values (k <= n)
+        - Counter(arr) scans the list once and each insertion/update is O(1)
+          on average: O(n)
+        - count.values() iterates over k counts: O(k)
+        - set(...) hashes each of the k counts: O(k)
+        - len(...) and the comparison are O(1)
+        - Total: O(n) + O(k) + O(k) = O(n), since k <= n
+        - Worst case: hash collisions could degrade dict/set operations, but
+          this is not a practical concern for small integers
+        - Tighter bound: the counts sum to n, so there are at most O(sqrt(n))
+          distinct counts if they are all unique; this does not change the
+          overall O(n) cost dominated by counting
+
+        Space complexity: O(k)
+        - Counter stores one entry per distinct value: O(k)
+        - The set of counts holds at most k values (usually far fewer): O(k)
+        - No recursion, so no call stack overhead
+        - Total auxiliary space: O(k), which is O(n) in the worst case when
+          all values are distinct; the output is a single bool: O(1)
+        """
+        count = Counter(arr)
+        return len(set(count.values())) == len(count.keys())

problems/easy/unique_number_of_occurrences_1207/test_solution.py

@@ -0,0 +1,15 @@
+import pytest
+from tests.base_test import BaseTestSolution
+from .solution import Solution
+
+
+class TestSolution(BaseTestSolution):
+    solution = Solution()
+
+    @pytest.mark.parametrize("method_name, arr, expected, timeout", [
+        ('unique_occurrences', [1, 2, 2, 1, 1, 3], True, BaseTestSolution.DEFAULT_TIMEOUT),
+        ('unique_occurrences', [1, 2], False, BaseTestSolution.DEFAULT_TIMEOUT),
+        ('unique_occurrences', [-3, 0, 1, -3, 1, 1, 1, -3, 10, 0], True, BaseTestSolution.DEFAULT_TIMEOUT),
+    ])
+    def test_unique_occurrences(self, method_name, arr, expected, timeout):
+        self._run_test(self.solution, method_name, (arr,), expected, timeout)