def test_count_ones_iter(self):
# 8 -> 1000
self.assertEqual(1, count_ones_iter(8))
# 109 -> 1101101
self.assertEqual(5, count_ones_iter(109))
# 63 -> 111111
self.assertEqual(6, count_ones_iter(63))
# 0 -> 0
self.assertEqual(0, count_ones_iter(0))
def test_count_ones_iter(self):
# 8 -> 1000
self.assertEqual(1, count_ones_iter(8))
# 109 -> 1101101
self.assertEqual(5, count_ones_iter(109))
# 63 -> 111111
self.assertEqual(6, count_ones_iter(63))
# 0 -> 0
self.assertEqual(0, count_ones_iter(0))
""" Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight). For example, the 32-bit integer '11' has binary representation 00000000000000000000000000001011, so the function should return 3. T(n)- O(k) : k is the number of 1s present in binary representation. NOTE: this complexity is better than O(log n). e.g. for n = 00010100000000000000000000000000 only 2 iterations are required. Number of loops is equal to the number of 1s in the binary representation. """ from algorithms.bit import count_ones_iter, count_ones_recur n = 11 print(count_ones_iter(n)) print(count_ones_recur(n))