def test_count_one(self): s = SortedSet([1, 5, 7, 9]) self.assertEqual(s.count(7), 1)
def test_count_zero(self): s = SortedSet([1, 2, 3, 4, 5]) self.assertEqual(s.count(14), 0)
def test_count_zero(self): s = SortedSet([1, 5, 7, 9]) self.assertEqual(s.count(11), 0)
def test_count_one(self): s = SortedSet([1, 3, 4, 9]) self.assertEqual(s.count(9), 1)
def test_count_one(self): s = SortedSet([1, 2, 3, 4, 5]) self.assertEqual(s.count(4), 1)
def test_count_in(self): s = SortedSet([1, 2, 3, 4]) self.assertEqual(1, s.count(1))
def test_count_not_in(self): s = SortedSet([1, 2, 3, 4]) self.assertEqual(0, s.count(5))
def test_count_one(self): s = SortedSet([7, 14, 22, 23, 7]) self.assertEqual(s.count(7), 1)
def test_count_zero(self): s = SortedSet([7, 14, 22, 23, 7]) self.assertEqual(s.count(20), 0)
# 10_13-Improving Performance From O(N) to O(log n) from sorted_set import SortedSet from random import randrange s = SortedSet(randrange(1000) for _ in range(2000)) s len(s) [s.count(i) for i in range(1000)]