forked from TheAlgorithms/Python
-
Notifications
You must be signed in to change notification settings - Fork 0
/
palindrome.py
104 lines (83 loc) · 2.91 KB
/
palindrome.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
# Algorithms to determine if a string is palindrome
from timeit import timeit
test_data = {
"MALAYALAM": True,
"String": False,
"rotor": True,
"level": True,
"A": True,
"BB": True,
"ABC": False,
"amanaplanacanalpanama": True, # "a man a plan a canal panama"
}
# Ensure our test data is valid
assert all((key == key[::-1]) is value for key, value in test_data.items())
def is_palindrome(s: str) -> bool:
"""
Return True if s is a palindrome otherwise return False.
>>> all(is_palindrome(key) is value for key, value in test_data.items())
True
"""
start_i = 0
end_i = len(s) - 1
while start_i < end_i:
if s[start_i] == s[end_i]:
start_i += 1
end_i -= 1
else:
return False
return True
def is_palindrome_traversal(s: str) -> bool:
"""
Return True if s is a palindrome otherwise return False.
>>> all(is_palindrome_traversal(key) is value for key, value in test_data.items())
True
"""
end = len(s) // 2
n = len(s)
# We need to traverse till half of the length of string
# as we can get access of the i'th last element from
# i'th index.
# eg: [0,1,2,3,4,5] => 4th index can be accessed
# with the help of 1st index (i==n-i-1)
# where n is length of string
return all(s[i] == s[n - i - 1] for i in range(end))
def is_palindrome_recursive(s: str) -> bool:
"""
Return True if s is a palindrome otherwise return False.
>>> all(is_palindrome_recursive(key) is value for key, value in test_data.items())
True
"""
if len(s) <= 2:
return True
if s[0] == s[len(s) - 1]:
return is_palindrome_recursive(s[1:-1])
else:
return False
def is_palindrome_slice(s: str) -> bool:
"""
Return True if s is a palindrome otherwise return False.
>>> all(is_palindrome_slice(key) is value for key, value in test_data.items())
True
"""
return s == s[::-1]
def benchmark_function(name: str) -> None:
stmt = f"all({name}(key) is value for key, value in test_data.items())"
setup = f"from __main__ import test_data, {name}"
number = 500000
result = timeit(stmt=stmt, setup=setup, number=number)
print(f"{name:<35} finished {number:,} runs in {result:.5f} seconds")
if __name__ == "__main__":
for key, value in test_data.items():
assert is_palindrome(key) is is_palindrome_recursive(key)
assert is_palindrome(key) is is_palindrome_slice(key)
print(f"{key:21} {value}")
print("a man a plan a canal panama")
# finished 500,000 runs in 0.46793 seconds
benchmark_function("is_palindrome_slice")
# finished 500,000 runs in 0.85234 seconds
benchmark_function("is_palindrome")
# finished 500,000 runs in 1.32028 seconds
benchmark_function("is_palindrome_recursive")
# finished 500,000 runs in 2.08679 seconds
benchmark_function("is_palindrome_traversal")