Date: 18 July, 2023
Model: gpt-4
Temperature: 0.2
Passing runs: 4
Total runs: 10
Success rate: 40%
Tokens: 194
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: 2023-07-18 18:07:23
Duration: 22.43s
Completion tokens: 307
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'
'''
# Check if the first string ends with '(' and the second string starts with ')'
if lst[0].endswith('(') and lst[1].startswith(')'):
return 'Yes'
# Check if the first string starts with ')' and the second string ends with '('
elif lst[0].startswith(')') and lst[1].endswith('('):
return 'Yes'
# Check if both strings are balanced
elif lst[0].count('(') == lst[0].count(')') and lst[1].count('(') == lst[1].count(')'):
return 'Yes'
else:
return 'No'Start time: 2023-07-18 18:06:55
Duration: 27.76s
Completion tokens: 417
Result: Pass
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'
'''
# Concatenate the strings in both possible orders
str1 = lst[0] + lst[1]
str2 = lst[1] + lst[0]
# Check if either concatenated string is good
if is_good(str1) or is_good(str2):
return 'Yes'
else:
return 'No'
def is_good(s):
'''
Helper function to check if a string of parentheses is good.
A string is good if and only if all parentheses are balanced.
'''
# Initialize a counter for open parentheses
count = 0
# Iterate over the characters in the string
for char in s:
# If the character is an open parenthesis, increment the counter
if char == '(':
count += 1
# If the character is a close parenthesis, decrement the counter
elif char == ')':
count -= 1
# If the counter becomes negative at any point, the string is not good
if count < 0:
return False
# If the counter is zero at the end, the string is good
return count == 0Start time: 2023-07-18 18:06:28
Duration: 26.75s
Completion tokens: 255
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'
'''
# count the number of open and close parentheses in both strings
open_parens = sum(s.count('(') for s in lst)
close_parens = sum(s.count(')') for s in lst)
# if the number of open and close parentheses are equal, return 'Yes'
if open_parens == close_parens:
return 'Yes'
else:
return 'No'Start time: 2023-07-18 18:05:59
Duration: 29.23s
Completion tokens: 342
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'
'''
# Check if the first string has more open parentheses than close parentheses
if lst[0].count('(') >= lst[0].count(')'):
# If so, check if the second string has more close parentheses than open parentheses
if lst[1].count(')') >= lst[1].count('('):
return 'Yes'
# Check if the first string has more close parentheses than open parentheses
if lst[0].count(')') > lst[0].count('('):
# If so, check if the second string has more open parentheses than close parentheses
if lst[1].count('(') > lst[1].count(')'):
return 'Yes'
# If none of the above conditions are met, return 'No'
return 'No'Start time: 2023-07-18 18:05:30
Duration: 28.14s
Completion tokens: 356
Result: Pass
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'
'''
# sort the list so that the string with more open parentheses comes first
lst.sort(key=lambda s: s.count('('), reverse=True)
# concatenate the strings
s = ''.join(lst)
# keep track of the balance of the parentheses
balance = 0
# iterate over the characters in the string
for c in s:
# increase the balance for every open parenthesis
if c == '(':
balance += 1
# decrease the balance for every close parenthesis
else:
balance -= 1
# if the balance is negative at any point, the string is not good
if balance < 0:
return 'No'
# if the balance is not zero at the end, the string is not good
if balance != 0:
return 'No'
# otherwise, the string is good
return 'Yes'Start time: 2023-07-18 18:04:52
Duration: 37.82s
Completion tokens: 410
Result: Pass
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'
'''
# Concatenate the strings in both possible orders
str1 = lst[0] + lst[1]
str2 = lst[1] + lst[0]
# Check if either concatenated string is good
if is_good(str1) or is_good(str2):
return 'Yes'
else:
return 'No'
def is_good(s):
'''
Helper function to check if a string of parentheses is good.
'''
# Initialize a counter for open parentheses
open_parens = 0
# Iterate over the characters in the string
for char in s:
if char == '(':
# Increment the counter for an open parenthesis
open_parens += 1
else:
# Decrement the counter for a close parenthesis
open_parens -= 1
# If the counter goes negative at any point, the string is not good
if open_parens < 0:
return False
# If the counter is not zero at the end, the string is not good
return open_parens == 0Start time: 2023-07-18 18:04:26
Duration: 26.24s
Completion tokens: 260
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'
'''
# Check if the first string ends with '(' and the second string starts with ')'
if lst[0][-1] == '(' and lst[1][0] == ')':
return 'Yes'
# Check if the second string ends with '(' and the first string starts with ')'
elif lst[1][-1] == '(' and lst[0][0] == ')':
return 'Yes'
else:
return 'No'Start time: 2023-07-18 18:04:00
Duration: 26.22s
Completion tokens: 358
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'
'''
# Check if the first string ends with '(' and the second string starts with ')'
if lst[0].endswith('(') and lst[1].startswith(')'):
return 'Yes'
# Check if the second string ends with '(' and the first string starts with ')'
elif lst[1].endswith('(') and lst[0].startswith(')'):
return 'Yes'
# Check if both strings start and end with the same type of parentheses
elif lst[0].startswith('(') and lst[0].endswith(')') and lst[1].startswith('(') and lst[1].endswith(')'):
return 'Yes'
elif lst[0].startswith(')') and lst[0].endswith('(') and lst[1].startswith(')') and lst[1].endswith('('):
return 'Yes'
else:
return 'No'Start time: 2023-07-18 18:03:33
Duration: 26.85s
Completion tokens: 340
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'
'''
# Count the number of open and close parentheses in each string
open_parens = [s.count('(') for s in lst]
close_parens = [s.count(')') for s in lst]
# Check if the total number of open and close parentheses are equal
if sum(open_parens) != sum(close_parens):
return 'No'
# Check if it's possible to order the strings so that no close parenthesis
# comes before an open parenthesis
if max(close_parens[0] - open_parens[1], 0) > open_parens[0] or max(close_parens[1] - open_parens[0], 0) > open_parens[1]:
return 'No'
return 'Yes'Start time: 2023-07-18 18:03:15
Duration: 17.74s
Completion tokens: 247
Result: Pass
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'
'''
for i in range(2):
s = lst[i] + lst[1 - i]
balance = 0
for c in s:
if c == '(':
balance += 1
else:
balance -= 1
if balance < 0:
break
if balance == 0:
return 'Yes'
return 'No'