Tasks Details
easy
1.
Brackets
Determine whether a given string of parentheses (multiple types) is properly nested.
Task Score
100%
Correctness
100%
Performance
100%
A string S consisting of N characters is considered to be properly nested if any of the following conditions is true:
- S is empty;
- S has the form "(U)" or "[U]" or "{U}" where U is a properly nested string;
- S has the form "VW" where V and W are properly nested strings.
For example, the string "{[()()]}" is properly nested but "([)()]" is not.
Write a function:
def solution(S)
that, given a string S consisting of N characters, returns 1 if 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..200,000];
- string S is made only of the following characters: '(', '{', '[', ']', '}' and/or ')'.
Copyright 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Python
Time spent on task 1 minutes
Notes
not defined yet
Task timeline
Code: 02:41:52 UTC,
java,
autosave
Code: 02:42:02 UTC,
py,
autosave
def solution(S):
S_length = len(S)
if S_length == 0:
return 0
temp_stack = []
for bracket in S:
if bracket == '(' or bracket == '{' or bracket == '[':
temp_stack.append(bracket)
elif len(temp_stack) == 0:
return 0
if bracket == ')':
if temp_stack.pop() != '(':
return 0
if bracket == '}':
if temp_stack.pop() != '{':
return 0
if bracket == ']':
if temp_stack.pop() != '[':
return 0
if len(temp_stack) == 0:
return 1
else:
return 0
Code: 02:42:08 UTC,
py,
verify,
result: Passed
def solution(S):
temp_stack = []
for bracket in S:
if bracket == '(' or bracket == '{' or bracket == '[':
temp_stack.append(bracket)
elif len(temp_stack) == 0:
return 0
if bracket == ')':
if temp_stack.pop() != '(':
return 0
if bracket == '}':
if temp_stack.pop() != '{':
return 0
if bracket == ']':
if temp_stack.pop() != '[':
return 0
if len(temp_stack) == 0:
return 1
else:
return 0
Analysis
Code: 02:42:11 UTC,
py,
verify,
result: Passed
def solution(S):
temp_stack = []
for bracket in S:
if bracket == '(' or bracket == '{' or bracket == '[':
temp_stack.append(bracket)
elif len(temp_stack) == 0:
return 0
if bracket == ')':
if temp_stack.pop() != '(':
return 0
if bracket == '}':
if temp_stack.pop() != '{':
return 0
if bracket == ']':
if temp_stack.pop() != '[':
return 0
if len(temp_stack) == 0:
return 1
else:
return 0
Analysis
Code: 02:42:14 UTC,
py,
final,
score:
100
def solution(S):
temp_stack = []
for bracket in S:
if bracket == '(' or bracket == '{' or bracket == '[':
temp_stack.append(bracket)
elif len(temp_stack) == 0:
return 0
if bracket == ')':
if temp_stack.pop() != '(':
return 0
if bracket == '}':
if temp_stack.pop() != '{':
return 0
if bracket == ']':
if temp_stack.pop() != '[':
return 0
if len(temp_stack) == 0:
return 1
else:
return 0
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
4.
0.036 s
OK
5.
0.036 s
OK
1.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
4.
0.036 s
OK
5.
0.036 s
OK
expand all
Performance tests
1.
0.100 s
OK
2.
0.036 s
OK
3.
0.044 s
OK
1.
0.044 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
1.
0.100 s
OK
multiple_full_binary_trees
sequence of full trees of the form T=(TT), depths [1..10..1], with/without some brackets at the end, length=49K+
sequence of full trees of the form T=(TT), depths [1..10..1], with/without some brackets at the end, length=49K+
✔
OK
1.
0.052 s
OK
2.
0.056 s
OK
3.
0.056 s
OK
4.
0.052 s
OK
5.
0.036 s
OK
broad_tree_with_deep_paths
string of the form [TTT...T] of 300 T's, each T being '{{{...}}}' nested 200-fold, length=120K+
string of the form [TTT...T] of 300 T's, each T being '{{{...}}}' nested 200-fold, length=120K+
✔
OK
1.
0.080 s
OK
2.
0.080 s
OK