Given a string that contains only the following => ‘{‘, ‘}’, ‘(‘, ‘)’, ‘[’, ‘]’. At some places there is ‘X’ in place of any bracket. Determine whether by replacing all ‘X’s with appropriate bracket, is it possible to make a valid bracket sequence.
Examples:
Input : S = "{(X[X])}"
Output : Balanced
Input : S = "[{X}(X)]"
Output : Not balanced
I tried to work it out like this, and it works for examples above. But it doesn't work for all examples eg. (it should be balanced but it says it's not)
Input: S = "([X}])"
Output: Not balanced
I tried to work it out but i can't find a solution. Please help.
class Stack:
def __init__(self):
self.data = []
def insert(self, x):
self.data.append(x)
def empty(self):
return len(self.data) == 0
def remove(self):
if self.empty():
raise ValueError('Stack is empty.')
self.data.pop()
def top_element(self):
if self.empty():
raise ValueError('Stack is empty.')
return self.data[-1]
def is_matching(a, b):
if a == "(" and b == ")":
return True
elif a == "[" and b == "]":
return True
elif a == "{" and b == "}":
return True
elif a == "X":
return True
return False
def is_balanced(expression,elements=Stack(),ind=0):
if ind == len(expression):
return elements.empty()
pre_brackets = "([{"
post_brackets = ")]}"
char = expression[ind]
if char in pre_brackets:
elements.insert(char)
return is_balanced(expression,elements,ind+1)
elif char in post_brackets:
if elements.empty() :
return False
if not is_matching(elements.top_element(), char):
return False
elements.remove()
return is_balanced(expression,elements,ind+1)
elif char == "X":
temp = Stack()
temp.insert(char)
result = (is_balanced(expression,temp,ind+1))
if result:
return True
if elements.empty():
return False
elements.remove()
return is_balanced(expression,elements,ind+1)
expression = "([X}])"
if expression == "":
print("No brackets in expression!")
elif len(expression) % 2 != 0:
print("Not balanced")
elif is_balanced(expression):
print("Balanced")
else:
print("Not Balanced")
You can do it by recursively testing all possible replacements for an X:
def can_be_balanced(expr):
pairs = "{}[]()"
opening_brackets = pairs[::2]
closing_brackets = pairs[1::2]
closer = {o:c for o, c in zip(opening_brackets, closing_brackets)}
opener = {c:o for o, c in zip(opening_brackets, closing_brackets)}
stack = []
for item in expr:
if item in opening_brackets:
# we append opening brackets to the stack
stack.append(item)
elif item in closing_brackets:
if not stack or stack[-1] != opener[item]:
# the closing bracket doesn't match the top of the stack
return False
else:
# if it does, we remove the matching opening bracket
stack.pop()
elif item == 'X':
# X could be any of the opening brackets,
possible = list(opening_brackets)
if stack and stack[-1] in opening_brackets:
# or the closing bracket matching the top of the stack
possible.append(closer[stack[-1]])
for pos in possible:
# we replace this X, the first one remaining in expr
test_expr = expr.replace('X', pos, 1)
if can_be_balanced(test_expr):
# This is just in order to print the working solution we just found,
# you may remove these two lines
if not 'X' in test_expr:
print(test_expr)
return True
# None of the replacements for X gave a balanced expression
return False
else:
raise ValueError(f'Invalid item {item} in {expr}')
# The expression is balanced if and only if the stack ends up empty
return not stack
Testing on your sample data:
tests = [("[{X}(X)]", False),
("{(X[X])}", True),
("([X}])", True),
]
for test in tests:
print(test[0], ': should be', test[1])
print(can_be_balanced(test[0]))
print('-'*20)
correctly outputs (with the balanced expression in case it can be done):
[{X}(X)] : should be False
False
--------------------
{(X[X])} : should be True
{([[]])}
True
--------------------
([X}]) : should be True
([{}])
True
--------------------
Note that a major problem in your code is that you only test the end of the expression, starting at the position of the X. Beware also of the mutable default argument elements=Stack() that would leave you with the remnants of the previous call to the function instead of a fresh, empty Stack object.
Related
I wrote a regex code which compares two strings. It recognises a special character '?' that allows zero or more instances of previous character. It works fine until there are two or more occasions of '?' in the string. And I can't make out why.
def single_character_string(a, b) -> "return True if characters match":
"""check if two characters match"""
if len(a) == 0:
return True
elif len(b) == 0:
return False
else:
if a == '.':
return True
else:
if a == b:
return True
else:
return False
def meta_question_result(temp):
if len(temp) >= 2:
if temp[1] == '?':
k_1 = temp.replace(temp[0: 2], '') # no char
k_2 = temp.replace(temp[1], '') # char
return k_1, k_2
def check_pair_by_pair(template, check_string) -> "Strings are of Equal length! " \
"return True if lines are identical":
"""check if two strings match symbol by symbol. template may be less than string, the opposite
is False"""
if not template: # exit from recursion
return True
if not check_string: # exit from recursion
return False
if meta_question_result(template):
t_1, t_2 = meta_question_result(template)
if single_character_string(t_1[0], check_string[0]):
return check_pair_by_pair(t_1[1:], check_string[1:])
if single_character_string(t_2[0], check_string[0]):
return check_pair_by_pair(t_2[1:], check_string[1:])
else:
return False
elif single_character_string(template[0], check_string[0]):
return check_pair_by_pair(template[1:], check_string[1:])
else:
return False
reg, st = input().split("|")
print(check_pair_by_pair(reg, st))
reg = "co?lou?r"
st = "colour"
gives True as expected,
reg = "co?lou?r"
st = "clor"
gives True as expected,
but...
reg = "co?lou?r"
st = "color"
gives False. I expected True.
Found the bag.
Replace method replaces all instances of '?'. So the second '?' was replaced also and program didn't see it.
I should add an argument 'count' to replace method that is equal to 1.
k_1 = temp.replace(temp[0: 2], '', 1) # no char
I need to write a function that given a string with parenthesis and/or square brackets it is able to evaluate if they appear in the correct order. For example, in this string '([b])(aa)' you can see that every time a parenthesis or square bracket is open, it is closed in the correct position. However, a string like '[(a])' it is not closing the parenthesis or square brackets in the correct order as it should be '[(a)]'.
The function should return True or False depending on this correct position of both elements. I have tried the following code, but this logic seems to be infinite and it is not working if I have more than two parenthesis or square brackets opened.
def parenthesis(string):
for a in range(len(string)):
if string[a] == "(":
for b in range(a,len(string)):
if string[b] == "[":
for c in range(b,len(string)):
if string[c] == "]":
for d in range(c,len(string)):
if string[d] == ")":
return True
elif string[b] == ")":
return True
else:
return False
If I run the function over the string "([b])(aa)" it is returning false as output.
parenthesis("([b])(aa)")
How can I rewrite this function so it evaluates all the parenthesis and square brackets combinations properly?
If a right parenthesis is open before a left, you got -1 and return False
def is_balanced(string):
cnt = 0
for char in string:
if char == '(': cnt += 1
if char == ')': cnt -= 1
if cnt < 0: return False
return True if cnt == 0 else False
This is one of the stack implementations I know:
def is_balanced(s):
stack = []
for char in s:
if char == "(" or char == "{" or char == "[":
stack.append(char)
elif len(stack) <= 0:
return False
elif char == ")" and stack.pop() != "(":
return False
elif char == "]" and stack.pop() != "[":
return False
elif char == "}" and stack.pop() != "{":
return False
if len(stack) == 0:
return True
return False
This version is more DRY than the prior answer:
def is_balanced(parens: str) -> bool:
# Link: https://stackoverflow.com/a/73341167/
parens_map ={'(':')','{':'}','[':']'}
stack = []
for paren in parens:
if paren in parens_map: # is open
stack.append(paren)
elif paren in parens_map.values(): # is close
if (not stack) or (paren != parens_map[stack.pop()]):
return False
return not stack
I am trying to come up with a code for the valid parentheses problem (https://leetcode.com/problems/valid-parentheses/) but I am having trouble understanding the logic of this problem.
This is my pseudocode
Iterate over characters in the string.
Get the current element.
Check if current element is == next element.
if it is, return true, else return false.
This was my solution but it doesn't work.
class Solution(object):
def isValid(self,s):
# iterate through the index in the string
for i in range(0,len(s)-1)):
# if the current index is equal to the next index, return true
if(s[i] == s[i+1]):
return True
I asked this question earlier and got the feedback that my code looking for places where the same symbol appears twice in a way. Whereas, I want my code to match up pairs of brackets.
Could someone help me understand the logic of this problem?
Here's how I would approach it:
class Solution(object):
def isValid(self, s):
stack = []
for char in s:
if "(" == char:
stack.append(")")
elif "{" == char:
stack.append("}")
elif "[" == char:
stack.append("]")
elif not stack or stack.pop() != char:
return False
return not stack
def test_is_valid(s, expected):
result = is_valid(s)
print(repr(s), result)
assert result == expected
is_valid = Solution().isValid
test_is_valid("()", True)
test_is_valid("()[]{}", True)
test_is_valid("(]", False)
test_is_valid("", True)
test_is_valid("{", False)
test_is_valid("}", False)
test_is_valid("[{([])}]", True)
Output:
'()' True
'()[]{}' True
'(]' False
'' True
'{' False
'}' False
'[{([])}]' True
This is my first time asking a question on here, and I am only doing this because I have spent the past week trying to figure this out and haven't been able to. I found similar questions, but the results did not help me. I have to take an infix expression and calculate the results using two stacks, one for the operator and one for the numbers. An example would be 6 - ( 5 - 3 ) * ( 4 + 2 ) = -6 or 3 * 11 / 8 + 5 – 4 * 7 = -18.875. I just cannot figure out how to get this to work. Currently my code is this:
class NumStack:
def __init__(self):
"""Create an empty stack."""
self._data = [] #nonpublic list instance
def __len__(self):
"""Return the number of elements in the stack."""
return len(self._data)
def is_empty(self):
"""Return True if the stack is empty."""
return len(self._data) == 0
def push(self,e):
"""Add element e to the top of the stack."""
self._data.append(e) #new item stored at end of list
print(self._data)
def top(self):
"""Return (but do not remove) the element at the top of the stack.
Raise Empty exception if the stack is empty"""
if self.is_empty():
return
return self._data[-1] #the last item in the list
def pop(self):
"""Remove and return the element from the top of the stack (i.e, LIFO)
Raise Empty exception if the stack is empty."""
if self.is_empty():
return "empty"
return self._data.pop() #remove last item from list
def str(self):
return self._data
class OperatorStack:
def __init__(self):
"""Create an empty stack."""
self._data = [] #nonpublic list instance
def __len__(self):
"""Return the number of elements in the stack."""
return len(self._data)
def is_empty(self):
"""Return True if the stack is empty."""
length = len(self._data)
if length == 0:
return True
else:
return False
def push(self,e):
"""Add element e to the top of the stack."""
self._data.append(e) #new item stored at end of list
print(self._data)
def top(self):
"""Return (but do not remove) the element at the top of the stack.
Raise Empty exception if the stack is empty"""
if self.is_empty():
return
return self._data[-1] #the last item in the list
def pop(self):
"""Remove and return the element from the top of the stack (i.e, LIFO)
Raise Empty exception if the stack is empty."""
length = len(self)
if length == 0:
print("list is empty")
else:
if self.is_empty():
return
return self._data.pop()
def str(self):
return self._data
def main():
expression = str(input("Enter an expression: "))
expression = expression.split()
print(expression)
N = NumStack()
O = OperatorStack()
new = []
NewOP = []
NewNum = [0,0]
for e in expression:
if e == '(' or e == ')' or e == '+' or e == '-' or e == '*' or e == '/':
O.push(e)
else:
N.push(e)
while O.is_empty() == False or N.is_empty() == False:
TmpOp = O.top()
if TmpOp == ')':
O.pop()
elif TmpOp == '(':
O.pop()
if TmpOp != '(' and TmpOp != ')':
new.append(N.pop())
new.append(O.pop())
print(TmpOp)
while TmpOp == ')':
if N.top() != "empty":
NewNum[1] = N.pop()
if N.top() != "empty":
NewNum[0] = N.top()
print(NewNum[0],NewNum[1])
if O.pop() == '+':
num = float(NewNum[1]) + float(NewNum[0])
new.append(num)
print(num)
O.pop()
break
elif O.pop() == '-':
num = float(NewNum[0]) - float(NewNum[1])
new.append(num)
print(num)
O.pop()
break
elif O.pop() == '*':
num = NewNum[1]*NewNum[0]
new.append(num)
print(num)
# O.pop()
break
elif O.pop() == '/':
num = NewNum[1]/NewNum[0]
new.append(num)
print(num)
# O.pop()
break
if O.top() == ')':
O.pop()
break
if O.__len__() == 0 and N.__len__() == 0:
break
continue
while TmpOp != ')' and TmpOp != '(':
new.append(N.pop())
new.append(O.pop())
print(new)
if O.__len__() == 0 and N.__len__() == 0:
break
print(new)
main()
I believe my classes to be correct, I just cannot find a way of extracting the needed information correctly. I am trying to pop the items into a list and when I get to a parenthesis I go ahead and perform the calculation. If I could just get the correct numbers into my list then I know I can get it to work. I just need to get the "new" to contain the correct numbers. With the problem: "6 - ( 5 - 3 ) * ( 4 + 2 )" I should get [6.0, '*', 2.0, '-', 6]. It is not coming out to that. I really appreciate any help that can be given.
I am trying to return one instead of true in python.
The code i am working on is:
delimiters = ( '()', '[]', '{}', "''", '""' )
esc = '\\'
def is_balanced(s, delimiters=delimiters, esc=esc):
stack = []
opening = tuple(str[0] for str in delimiters)
closing = tuple(str[1] for str in delimiters)
for i, c in enumerate(s):
if len(stack) and stack[-1] == -1:
stack.pop()
elif c in esc:
stack.append(-1)
elif c in opening and (not len(stack) or opening[stack[-1]] != closing[stack[-1]]):
stack.append(opening.index(c))
elif c in closing:
if len(stack) == 0 or closing.index(c) != stack[-1]:
return False
stack.pop()
return len(stack) == 0
num_cases = raw_input()
num_cases = int(num_cases)
for num in range(num_cases):
s = raw_input()
print is_balanced(s)
It basically checks whether the string typed is balanced or not. If balanced, should return 1 and if not 0.
I tried this:
1
Test string
True
It returns true. I would like it to return 1. How do i do it?
Alternatively you could cast your boolean to an int:
>>>myBoolean = True
>>>int(myBoolean)
1
>>>myBoolean = False
>>>int(myBoolean)
0
Huh? You change the code:
Instead of
return False
write
return 0
and instead of
return len(stack) == 0
write
if len(stack) == 0:
return 1
return 0
The latter 3-liner can be rewritten on a single line, but I chose the above for clarity.
return 1 if len(stack) == 0 else 0
This concisely changes the return value of is_balanced, and is equivalent to:
if len(stack) == 0:
return 1
else:
return 0
Of course you could keep is_balanced unchanged and print (in similar notation):
1 if is_balanced(s) else 0
Just use
print +is_balanced(s)
instead.