feat: [LeetCode #933] Number Of Recent Calls

Тип: Queue
Сложность: easy
Временная сложность: O(1)
Пространственная сложность: O(n)

-

Ссылка: https://leetcode.com/problems/number-of-recent-calls/

Author: onixlas

src/queue/__init__.py

@@ -0,0 +1,17 @@
+from collections import deque
+
+
+class RecentCounter:
+    def __init__(self):
+        self.deque = deque([])
+        self._INTERVAL = 3000
+
+    def ping(self, t: int) -> int:
+        previous = t - self._INTERVAL
+
+        while self.deque and self.deque[0] < previous:
+            self.deque.popleft()
+
+        self.deque.append(t)
+
+        return len(self.deque)

src/queue/number_of_recent_calls/__init__.py

@@ -0,0 +1,28 @@
+import pytest
+from src.queue.number_of_recent_calls.solution import RecentCounter
+
+
+@pytest.mark.parametrize(
+    "calls, args, expected",
+    [
+        (
+            ["RecentCounter", "ping", "ping", "ping", "ping"],
+            [[], [1], [100], [3001], [3002]],
+            [None, 1, 2, 3, 3],
+        ),
+    ],
+)
+def test_recent_counter(calls, args, expected):
+    obj = None
+    results = []
+
+    for call, arg, exp in zip(calls, args, expected):
+        if call == "RecentCounter":
+            obj = RecentCounter()
+            results.append(None)
+        else:
+            method = getattr(obj, call)
+            res = method(*arg)
+            results.append(res)
+
+    assert results == expected