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 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Python
Time spent on task 9 minutes
Notes
not defined yet
Code: 07:51:55 UTC,
java,
autosave
Code: 07:57:50 UTC,
py,
verify,
result: Passed
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(S):
cnt = 0
for s in S:
if s == '(': cnt = cnt + 1
elif s == ')': cnt = cnt - 1
if cnt == 0: return 1
else: return 0
User test case 1:
['(()())']
Analysis
Code: 07:59:29 UTC,
py,
verify,
result: Passed
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(S):
cnt = 0
for s in S:
if s == '(': cnt = cnt + 1
elif s == ')': cnt = cnt - 1
if cnt == 0: return 1
else: return 0
User test case 1:
['(()())']
User test case 2:
['(()))((())']
Analysis
Code: 07:59:52 UTC,
py,
verify,
result: Passed
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(S):
cnt = 0
for s in S:
if s == '(': cnt = cnt + 1
elif s == ')': cnt = cnt - 1
if cnt < 0: print('bad')
if cnt == 0: return 1
else: return 0
User test case 1:
['(()())']
User test case 2:
['(()))((())']
Analysis
Code: 08:00:04 UTC,
py,
verify,
result: Passed
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(S):
cnt = 0
for s in S:
if s == '(': cnt = cnt + 1
elif s == ')': cnt = cnt - 1
if cnt < 0: return 0
if cnt == 0: return 1
else: return 0
User test case 1:
['(()())']
User test case 2:
['(()))((())']
Analysis
Code: 08:00:14 UTC,
py,
verify,
result: Passed
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(S):
cnt = 0
for s in S:
if s == '(': cnt = cnt + 1
elif s == ')': cnt = cnt - 1
if cnt < 0: return 0
if cnt == 0: return 1
else: return 0
User test case 1:
['(()())']
User test case 2:
['(()))((())']
Analysis
Code: 08:00:16 UTC,
py,
final,
score:
100
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
1.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
expand all
Performance tests
1.
0.040 s
OK
2.
0.040 s
OK
3.
0.036 s
OK
1.
0.060 s
OK
2.
0.036 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.
0.044 s
OK
2.
0.044 s
OK
3.
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=1 million
string of the form (TTT...T) of 300 T's, each T being '(((...)))' nested 200-fold, length=1 million
✔
OK
1.
0.152 s
OK
2.
0.036 s
OK