Tasks Details
easy
1.
Nesting
Determine whether a given string of parentheses (single type) is properly nested.
Task Score
100%
Correctness
100%
Performance
100%
A string S consisting of N characters is called properly nested if:
- S is empty;
- S has the form "(U)" where U is a properly nested string;
- S has the form "VW" where V and W are properly nested strings.
For example, string "(()(())())" is properly nested but string "())" isn't.
Write a function:
def solution(S)
that, given a string S consisting of N characters, returns 1 if string S is properly nested and 0 otherwise.
For example, given S = "(()(())())", the function should return 1 and given S = "())", the function should return 0, as explained above.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [0..1,000,000];
- string S is made only of the characters '(' and/or ')'.
Copyright 2009–2024 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Python
Total time used 8 minutes
Effective time used 8 minutes
Notes
not defined yet
Task timeline
Code: 14:03:53 UTC,
py,
verify,
result: Failed
# you can write to stdout for debugging purposes, e.g.
# print "this is a debug message"
def solution(S):
# write your code in Python 2.7
stack = []
for idx in xrange(len(S)):
if S[idx]=='(':
stack.append(S[idx])
else:
if len(stack)==0:
return 0
else:
stack.pop()
return len(stack)==0?0:1
Analysis
expand all
Example tests
1.
0.061 s
RUNTIME ERROR,
tested program terminated unexpectedly
stderr:
Traceback (most recent call last): File "user.py", line 103, in <module> main() File "user.py", line 63, in main sol = __import__('solution') File "/tmp/solution.py", line 15 return len(stack)==0?0:1 ^ SyntaxError: invalid syntax
1.
0.060 s
RUNTIME ERROR,
tested program terminated unexpectedly
stderr:
Traceback (most recent call last): File "user.py", line 103, in <module> main() File "user.py", line 63, in main sol = __import__('solution') File "/tmp/solution.py", line 15 return len(stack)==0?0:1 ^ SyntaxError: invalid syntax
Code: 14:04:04 UTC,
py,
verify,
result: Failed
# you can write to stdout for debugging purposes, e.g.
# print "this is a debug message"
def solution(S):
# write your code in Python 2.7
stack = []
for idx in xrange(len(S)):
if S[idx]=='(':
stack.append(S[idx])
else:
if len(stack)==0:
return 0
else:
stack.pop()
return len(stack)==0:0?1
Analysis
expand all
Example tests
1.
0.060 s
RUNTIME ERROR,
tested program terminated unexpectedly
stderr:
Traceback (most recent call last): File "user.py", line 103, in <module> main() File "user.py", line 63, in main sol = __import__('solution') File "/tmp/solution.py", line 15 return len(stack)==0:0?1 ^ SyntaxError: invalid syntax
1.
0.059 s
RUNTIME ERROR,
tested program terminated unexpectedly
stderr:
Traceback (most recent call last): File "user.py", line 103, in <module> main() File "user.py", line 63, in main sol = __import__('solution') File "/tmp/solution.py", line 15 return len(stack)==0:0?1 ^ SyntaxError: invalid syntax
Code: 14:06:32 UTC,
py,
verify,
result: Passed
# you can write to stdout for debugging purposes, e.g.
# print "this is a debug message"
def solution(S):
# write your code in Python 2.7
stack = []
for idx in xrange(len(S)):
if S[idx]=='(':
stack.append(S[idx])
else:
if len(stack)==0:
return 0
else:
stack.pop()
return 1 if len(stack)==0 else 0
Analysis
Code: 14:06:37 UTC,
py,
verify,
result: Passed
# you can write to stdout for debugging purposes, e.g.
# print "this is a debug message"
def solution(S):
# write your code in Python 2.7
stack = []
for idx in xrange(len(S)):
if S[idx]=='(':
stack.append(S[idx])
else:
if len(stack)==0:
return 0
else:
stack.pop()
return 1 if len(stack)==0 else 0
Analysis
Code: 14:06:39 UTC,
py,
final,
score: 
100
# you can write to stdout for debugging purposes, e.g.
# print "this is a debug message"
def solution(S):
# write your code in Python 2.7
stack = []
for idx in xrange(len(S)):
if S[idx]=='(':
stack.append(S[idx])
else:
if len(stack)==0:
return 0
else:
stack.pop()
return 1 if len(stack)==0 else 0
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
0.058 s
OK
2.
0.058 s
OK
1.
0.061 s
OK
1.
0.059 s
OK
2.
0.060 s
OK
3.
0.058 s
OK
1.
0.062 s
OK
2.
0.062 s
OK
3.
0.058 s
OK
expand all
Performance tests
1.
0.061 s
OK
2.
0.062 s
OK
3.
0.059 s
OK
1.
0.091 s
OK
2.
0.058 s
OK
multiple_full_binary_trees
sequence of full trees of the form T=(TT), depths [1..10..1], with/without unmatched ')' at the end, length=49K+
sequence of full trees of the form T=(TT), depths [1..10..1], with/without unmatched ')' at the end, length=49K+
✔
OK
1.
0.067 s
OK
2.
0.070 s
OK
3.
0.059 s
OK
broad_tree_with_deep_paths
string of the form (TTT...T) of 300 T's, each T being '(((...)))' nested 200-fold, length=1 million
string of the form (TTT...T) of 300 T's, each T being '(((...)))' nested 200-fold, length=1 million
✔
OK
1.
0.200 s
OK
2.
0.061 s
OK