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 2 minutes
Notes
not defined yet
Code: 09:13:58 UTC,
java,
autosave
Code: 09:14:09 UTC,
py,
autosave
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(S):
cnt = [0] * 3
stack = []
for s in S:
if s == '(':
cnt[0] = cnt[0] + 1
stack.append(0)
elif s == ')':
if stack[-1] != 0: return 0
stack.pop()
cnt[0] = cnt[0] - 1
elif s == '[':
stack.append(1)
cnt[1] = cnt[1] + 1
elif s == ']':
if stack[-1] != 1: return 0
stack.pop()
cnt[1] = cnt[1] - 1
elif s == '{':
stack.append(2)
cnt[2] = cnt[2] + 1
elif s == '}':
if stack[-1] != 2: return 0
stack.pop()
cnt[2] = cnt[2] - 1
if any(v < 0 for v in cnt): return 0
if all(v == 0 for v in cnt) and len(stack) == 0:
return 1
else:
return 0
Code: 09:14:19 UTC,
py,
autosave
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(S):
cnt = [0] * 3
stack = []
for s in S:
if s == '(':
cnt[0] = cnt[0] + 1
stack.append(0)
elif s == ')':
if stack[-1] != 0: return 0
stack.pop()
cnt[0] = cnt[0] - 1
elif s == '[':
stack.append(1)
cnt[1] = cnt[1] + 1
elif s == ']':
if stack[-1] != 1: return 0
stack.pop()
cnt[1] = cnt[1] - 1
elif s == '{':
stack.append(2)
cnt[2] = cnt[2] + 1
elif s == '}':
if stack[-1] != 2: return 0
stack.pop()
cnt[2] = cnt[2] - 1
if any(v < 0 for v in cnt): return 0
if all(v == 0 for v in cnt) or len(stack) == 0:
return 1
else:
return 0
Code: 09:14:30 UTC,
py,
autosave
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(S):
cnt = [0] * 3
stack = []
for s in S:
if s == '(':
cnt[0] = cnt[0] + 1
stack.append(0)
elif s == ')':
if stack[-1] != 0: return 0
stack.pop()
cnt[0] = cnt[0] - 1
elif s == '[':
stack.append(1)
cnt[1] = cnt[1] + 1
elif s == ']':
if stack[-1] != 1: return 0
stack.pop()
cnt[1] = cnt[1] - 1
elif s == '{':
stack.append(2)
cnt[2] = cnt[2] + 1
elif s == '}':
if stack[-1] != 2: return 0
stack.pop()
cnt[2] = cnt[2] - 1
if any(v < 0 for v in cnt): return 0
if all(v == 0 for v in cnt) and len(stack) == 0:
return 1
else:
return 0
Code: 09:14:47 UTC,
py,
autosave
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(S):
cnt = [0] * 3
stack = []
for s in S:
if s == '(':
cnt[0] = cnt[0] + 1
stack.append(0)
elif s == ')':
if len(stack) == 0 or stack[-1] != 0: return 0
stack.pop()
cnt[0] = cnt[0] - 1
elif s == '[':
stack.append(1)
cnt[1] = cnt[1] + 1
elif s == ']':
if stack[-1] != 1: return 0
stack.pop()
cnt[1] = cnt[1] - 1
elif s == '{':
stack.append(2)
cnt[2] = cnt[2] + 1
elif s == '}':
if stack[-1] != 2: return 0
stack.pop()
cnt[2] = cnt[2] - 1
if any(v < 0 for v in cnt): return 0
if all(v == 0 for v in cnt) and len(stack) == 0:
return 1
else:
return 0
Code: 09:15:03 UTC,
py,
autosave
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(S):
cnt = [0] * 3
stack = []
for s in S:
if s == '(':
cnt[0] = cnt[0] + 1
stack.append(0)
elif s == ')':
if len(stack) == 0 or stack[-1] != 0: return 0
stack.pop()
cnt[0] = cnt[0] - 1
elif s == '[':
stack.append(1)
cnt[1] = cnt[1] + 1
elif s == ']':
if len(stack) == 0 or stack[-1] != 1: return 0
stack.pop()
cnt[1] = cnt[1] - 1
elif s == '{':
stack.append(2)
cnt[2] = cnt[2] + 1
elif s == '}':
if len(stack) == 0 or stack[-1] != 2: return 0
stack.pop()
cnt[2] = cnt[2] - 1
if any(v < 0 for v in cnt): return 0
if all(v == 0 for v in cnt) and len(stack) == 0:
return 1
else:
return 0
Code: 09:15:36 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] * 3
stack = []
for s in S:
if s == '(':
cnt[0] = cnt[0] + 1
stack.append(0)
elif s == ')':
if len(stack) == 0 or stack[-1] != 0: return 0
stack.pop()
cnt[0] = cnt[0] - 1
elif s == '[':
stack.append(1)
cnt[1] = cnt[1] + 1
elif s == ']':
if len(stack) == 0 or stack[-1] != 1: return 0
stack.pop()
cnt[1] = cnt[1] - 1
elif s == '{':
stack.append(2)
cnt[2] = cnt[2] + 1
elif s == '}':
if len(stack) == 0 or stack[-1] != 2: return 0
stack.pop()
cnt[2] = cnt[2] - 1
if any(v < 0 for v in cnt): return 0
if all(v == 0 for v in cnt) and len(stack) == 0:
return 1
else:
return 0
Analysis
Code: 09:15:38 UTC,
py,
final,
score:
100
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(S):
cnt = [0] * 3
stack = []
for s in S:
if s == '(':
cnt[0] = cnt[0] + 1
stack.append(0)
elif s == ')':
if len(stack) == 0 or stack[-1] != 0: return 0
stack.pop()
cnt[0] = cnt[0] - 1
elif s == '[':
stack.append(1)
cnt[1] = cnt[1] + 1
elif s == ']':
if len(stack) == 0 or stack[-1] != 1: return 0
stack.pop()
cnt[1] = cnt[1] - 1
elif s == '{':
stack.append(2)
cnt[2] = cnt[2] + 1
elif s == '}':
if len(stack) == 0 or stack[-1] != 2: return 0
stack.pop()
cnt[2] = cnt[2] - 1
if any(v < 0 for v in cnt): return 0
if all(v == 0 for v in cnt) and len(stack) == 0:
return 1
else:
return 0Analysis 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.
0.248 s
OK
2.
0.036 s
OK
3.
0.052 s
OK
1.
0.056 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
1.
0.236 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.
0.088 s
OK
2.
0.092 s
OK
3.
0.088 s
OK
4.
0.088 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.
0.184 s
OK
2.
0.180 s
OK