Date: 01 September, 2023
Model: codellama-34b-instruct
Temperature: 0.2
Passing runs: 0
Total runs: 10
Success rate: 0%
def match_parens(lst):
'''
You are given a list of two strings, both strings consist of open
parentheses '(' or close parentheses ')' only.
Your job is to check if it is possible to concatenate the two strings in
some order, that the resulting string will be good.
A string S is considered to be good if and only if all parentheses in S
are balanced. For example: the string '(())()' is good, while the string
'())' is not.
Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
Examples:
match_parens(['()(', ')']) == 'Yes'
match_parens([')', ')']) == 'No'
'''The above is an incomplete Python code fragment. Return the complete and correct code with no additional text.
def match_parens(lst):
'''
You are given a list of two strings, both strings consist of open
parentheses '(' or close parentheses ')' only.
Your job is to check if it is possible to concatenate the two strings in
some order, that the resulting string will be good.
A string S is considered to be good if and only if all parentheses in S
are balanced. For example: the string '(())()' is good, while the string
'())' is not.
Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
Examples:
match_parens(['()(', ')']) == 'Yes'
match_parens([')', ')']) == 'No'
'''def check(s):
val = 0
for i in s:
if i == '(':
val = val + 1
else:
val = val - 1
if val < 0:
return False
return True if val == 0 else False
S1 = lst[0] + lst[1]
S2 = lst[1] + lst[0]
return 'Yes' if check(S1) or check(S2) else 'No'def check(candidate):
# Check some simple cases
assert candidate(['()(', ')']) == 'Yes'
assert candidate([')', ')']) == 'No'
assert candidate(['(()(())', '())())']) == 'No'
assert candidate([')())', '(()()(']) == 'Yes'
assert candidate(['(())))', '(()())((']) == 'Yes'
assert candidate(['()', '())']) == 'No'
assert candidate(['(()(', '()))()']) == 'Yes'
assert candidate(['((((', '((())']) == 'No'
assert candidate([')(()', '(()(']) == 'No'
assert candidate([')(', ')(']) == 'No'
# Check some edge cases that are easy to work out by hand.
assert candidate(['(', ')']) == 'Yes'
assert candidate([')', '(']) == 'Yes'Start time: 2023-09-01 02:46:37
Duration: 12.47s
Result: Fail
def match_parens(lst):
'''
You are given a list of two strings, both strings consist of open
parentheses '(' or close parentheses ')' only.
Your job is to check if it is possible to concatenate the two strings in
some order, that the resulting string will be good.
A string S is considered to be good if and only if all parentheses in S
are balanced. For example: the string '(())()' is good, while the string
'())' is not.
Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
Examples:
match_parens(['()(', ')']) == 'Yes'
match_parens([')', ')']) == 'No'
'''
left = 0
right = 0
for char in lst[0]:
if char == '(':
left += 1
else:
right += 1
for char in lst[1]:
if char == '(':
left += 1
else:
right += 1
if left == right:
return 'Yes'
else:
return 'No'Start time: 2023-09-01 02:49:35
Duration: 12.64s
Result: Fail
def match_parens(lst):
'''
You are given a list of two strings, both strings consist of open
parentheses '(' or close parentheses ')' only.
Your job is to check if it is possible to concatenate the two strings in
some order, that the resulting string will be good.
A string S is considered to be good if and only if all parentheses in S
are balanced. For example: the string '(())()' is good, while the string
'())' is not.
Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
Examples:
match_parens(['()(', ')']) == 'Yes'
match_parens([')', ')']) == 'No'
'''
left = 0
right = 0
for char in lst[0]:
if char == '(':
left += 1
else:
right += 1
for char in lst[1]:
if char == '(':
left += 1
else:
right += 1
if left == right:
return 'Yes'
else:
return 'No'Start time: 2023-09-01 02:52:33
Duration: 21.92s
Result: Fail
def match_parens(lst):
'''
You are given a list of two strings, both strings consist of open
parentheses '(' or close parentheses ')' only.
Your job is to check if it is possible to concatenate the two strings in
some order, that the resulting string will be good.
A string S is considered to be good if and only if all parentheses in S
are balanced. For example: the string '(())()' is good, while the string
'())' is not.
Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
Examples:
match_parens(['()(', ')']) == 'Yes'
match_parens([')', ')']) == 'No'
'''
# Initialize a counter to keep track of the number of open parentheses
count = 0
# Iterate through the first string
for char in lst[0]:
# If the current character is an open parenthesis, increment the counter
if char == '(':
count += 1
# If the current character is a close parenthesis, decrement the counter
elif char == ')':
count -= 1
# If the counter becomes negative, return 'No'
if count < 0:
return 'No'
# If the counter is not zero after iterating through the first string, return 'No'
if count != 0:
return 'No'
# Initialize a counter to keep track of the number of open parentheses
count = 0
# Iterate through the second string
for char in lst[1]:
# If the current character is an open parenthesis, increment the counter
if char == '(':
count += 1
# If the current character is a close parenthesis, decrement the counter
elif char == ')':
count -= 1
# If the counter becomes negative, return 'No'
if count < 0:
return 'No'
# If the counter is not zero after iterating through the second string, return 'No'
if count != 0:
return 'No'
# If the counter is zero after iterating through both strings, return 'Yes'
return 'Yes'Start time: 2023-09-01 02:55:26
Duration: 23.56s
Result: Fail
def match_parens(lst):
'''
You are given a list of two strings, both strings consist of open
parentheses '(' or close parentheses ')' only.
Your job is to check if it is possible to concatenate the two strings in
some order, that the resulting string will be good.
A string S is considered to be good if and only if all parentheses in S
are balanced. For example: the string '(())()' is good, while the string
'())' is not.
Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
Examples:
match_parens(['()(', ')']) == 'Yes'
match_parens([')', ')']) == 'No'
'''
# Initialize a variable to keep track of the number of open parentheses
open_parens = 0
# Iterate through the first string
for char in lst[0]:
# If the current character is an open parenthesis, increment the number of open parentheses
if char == '(':
open_parens += 1
# If the current character is a close parenthesis, decrement the number of open parentheses
elif char == ')':
open_parens -= 1
# If the number of open parentheses is ever negative, return 'No'
if open_parens < 0:
return 'No'
# If the number of open parentheses is not 0 after iterating through the first string, return 'No'
if open_parens != 0:
return 'No'
# Initialize a variable to keep track of the number of open parentheses
open_parens = 0
# Iterate through the second string
for char in lst[1]:
# If the current character is an open parenthesis, increment the number of open parentheses
if char == '(':
open_parens += 1
# If the current character is a close parenthesis, decrement the number of open parentheses
elif char == ')':
open_parens -= 1
# If the number of open parentheses is ever negative, return 'No'
if open_parens < 0:
return 'No'
# If the number of open parentheses is not 0 after iterating through the second string, return 'No'
if open_parens != 0:
return 'No'
# If both strings are good, return 'Yes'
return 'Yes'Start time: 2023-09-01 02:58:26
Duration: 12.21s
Result: Fail
def match_parens(lst):
'''
You are given a list of two strings, both strings consist of open
parentheses '(' or close parentheses ')' only.
Your job is to check if it is possible to concatenate the two strings in
some order, that the resulting string will be good.
A string S is considered to be good if and only if all parentheses in S
are balanced. For example: the string '(())()' is good, while the string
'())' is not.
Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
Examples:
match_parens(['()(', ')']) == 'Yes'
match_parens([')', ')']) == 'No'
'''
left_count = 0
right_count = 0
for char in lst[0]:
if char == '(':
left_count += 1
else:
right_count += 1
for char in lst[1]:
if char == '(':
left_count += 1
else:
right_count += 1
if left_count == right_count:
return 'Yes'
else:
return 'No'Start time: 2023-09-01 03:01:29
Duration: 12.31s
Result: Fail
def match_parens(lst):
'''
You are given a list of two strings, both strings consist of open
parentheses '(' or close parentheses ')' only.
Your job is to check if it is possible to concatenate the two strings in
some order, that the resulting string will be good.
A string S is considered to be good if and only if all parentheses in S
are balanced. For example: the string '(())()' is good, while the string
'())' is not.
Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
Examples:
match_parens(['()(', ')']) == 'Yes'
match_parens([')', ')']) == 'No'
'''
left_count = 0
right_count = 0
for char in lst[0]:
if char == '(':
left_count += 1
else:
right_count += 1
for char in lst[1]:
if char == '(':
left_count += 1
else:
right_count += 1
if left_count == right_count:
return 'Yes'
else:
return 'No'Start time: 2023-09-01 03:04:28
Duration: 12.52s
Result: Fail
def match_parens(lst):
'''
You are given a list of two strings, both strings consist of open
parentheses '(' or close parentheses ')' only.
Your job is to check if it is possible to concatenate the two strings in
some order, that the resulting string will be good.
A string S is considered to be good if and only if all parentheses in S
are balanced. For example: the string '(())()' is good, while the string
'())' is not.
Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
Examples:
match_parens(['()(', ')']) == 'Yes'
match_parens([')', ')']) == 'No'
'''
left = 0
right = 0
for char in lst[0]:
if char == '(':
left += 1
else:
right += 1
for char in lst[1]:
if char == '(':
left += 1
else:
right += 1
if left == right:
return 'Yes'
else:
return 'No'Start time: 2023-09-01 03:07:30
Duration: 12.66s
Result: Fail
def match_parens(lst):
'''
You are given a list of two strings, both strings consist of open
parentheses '(' or close parentheses ')' only.
Your job is to check if it is possible to concatenate the two strings in
some order, that the resulting string will be good.
A string S is considered to be good if and only if all parentheses in S
are balanced. For example: the string '(())()' is good, while the string
'())' is not.
Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
Examples:
match_parens(['()(', ')']) == 'Yes'
match_parens([')', ')']) == 'No'
'''
left_parens = 0
right_parens = 0
for char in lst[0]:
if char == '(':
left_parens += 1
else:
right_parens += 1
for char in lst[1]:
if char == '(':
left_parens += 1
else:
right_parens += 1
if left_parens == right_parens:
return 'Yes'
else:
return 'No'Start time: 2023-09-01 03:10:32
Duration: 12.04s
Result: Fail
def match_parens(lst):
'''
You are given a list of two strings, both strings consist of open
parentheses '(' or close parentheses ')' only.
Your job is to check if it is possible to concatenate the two strings in
some order, that the resulting string will be good.
A string S is considered to be good if and only if all parentheses in S
are balanced. For example: the string '(())()' is good, while the string
'())' is not.
Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
Examples:
match_parens(['()(', ')']) == 'Yes'
match_parens([')', ')']) == 'No'
'''
left = 0
right = 0
for char in lst[0]:
if char == '(':
left += 1
else:
right += 1
for char in lst[1]:
if char == '(':
left += 1
else:
right += 1
return 'Yes' if left == right else 'No'Start time: 2023-09-01 03:13:28
Duration: 12.79s
Result: Fail
def match_parens(lst):
'''
You are given a list of two strings, both strings consist of open
parentheses '(' or close parentheses ')' only.
Your job is to check if it is possible to concatenate the two strings in
some order, that the resulting string will be good.
A string S is considered to be good if and only if all parentheses in S
are balanced. For example: the string '(())()' is good, while the string
'())' is not.
Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
Examples:
match_parens(['()(', ')']) == 'Yes'
match_parens([')', ')']) == 'No'
'''
left = 0
right = 0
for char in lst[0]:
if char == '(':
left += 1
else:
right += 1
for char in lst[1]:
if char == '(':
left += 1
else:
right += 1
if left == right:
return 'Yes'
else:
return 'No'