We Keep Coding

How to check if the parentheses and brackets are balanced?

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

Runtime error when submitting Kattis problem in python: I can guess the data structure

I'm trying for a while now to submit my solution to the following problem on Kattis: I can guess the data structure!. However, I keep getting a runtime error; I cannot think of anywhere it can go wrong since it works with all my input. Here is my solution:
import heapq, sys
def throwin(q,s,h,x, results):
if results[0]:
q.append(x)
if results[1]:
s.append(x)
if results[2]:
heapq.heappush(h, -x)
return (q,s ,h )
def printstructures(l):
for j in l:
if j[0] and j[1] or j[0] and j[2] or j[1] and j[2]:
print("not sure")
elif not j[0] and not j[1] and not j[2]:
print("impossible")
elif j[0]:
print("queue")
elif j[1]:
print("stack")
else:
print("priority queue")
def main():
results_global = []
stackops = []
current = []
while True:
try:
line = input()
if len(line) == 1:
if len(current) != 0:
stackops.append(current)
current = []
else:
current.append(tuple(map(int, line.split(" "))))
except EOFError:
break
stackops.append(current)
for op in stackops:
q,s,h = [],[],[]
heapq._heapify_max(h)
results = [True, True, True]
for i in range(len(op)):
o, x = op[i]
if o == 1:
q,s,h = throwin(q,s,h,x, results)
else:
if len(q) == 0 or q[0] != x:
results[0] = False
else:
q.pop(0)
if len(s) == 0 or s[-1] != x:
results[1] = False
else:
s.pop()
if len(h) == 0 or h[0] != -x :
results[2] = False
else:
heapq.heappop(h)
if i == len(op)-1:
results_global.append(results)
printstructures(results_global)
if __name__ == "__main__":
main()
I was wondering if anyone can give me a push in the right direction and point out where my thinking is wrong or if I made a mistake somewhere I overlooked.

I had the same runtime-error problem for this question, I think it has something to do with python input/output EOFError. I couldn't figure the specific error out but I just put a try/except pass over my entire program and kattis accepted the solution.
import sys
try:
def solve(n):
stack = []
queue = []
priority_queue = []
type_ds = [True, True, True]
for i in range(n):
command, element = map(int, input().split())
if command == 1:
if type_ds[0] != False:
stack.append(element)
if type_ds[1] != False:
queue.append(element)
if type_ds[2] != False:
priority_queue.append(element)
elif command == 2:
if type_ds[0] != False:
if len(stack) == 0:
return "impossible"
if element != stack.pop():
type_ds[0] = False
stack.clear()
if type_ds[1] != False:
if len(queue) == 0:
return "impossible"
if element != queue.pop(0):
type_ds[1] = False
queue.clear()
if type_ds[2] != False:
if len(priority_queue) == 0:
return "impossible"
priority_queue.sort(reverse=True)
if element != priority_queue.pop(0):
type_ds[2] = False
priority_queue.clear()
if type_ds.count(True) > 1:
return "not sure"
elif type_ds.count(True) == 0:
return "impossible"
else:
if type_ds[0] == True:
return "stack"
elif type_ds[1] == True:
return "queue"
else:
return "priority queue"
for line in sys.stdin:
if line.strip() == "":
break
n = int(line.strip())
print(solve(n))
except:
pass

Are you sure you're breaking out of the while loop? Is it running correctly on your computer? In competitive programming, with time limits, it's normally not a good idea to use try/except statements, it's slowing down the script by a lot.

Infix evaluation in Python

I am trying to convert the code here http://www.geeksforgeeks.org/expression-evaluation/ to python. However, I am running into some trouble and can't figure out.
class evaluateString:
def evalString(self,expression):
valueStack = []
opStack = []
i=0
while(i<len(expression)):
if(expression[i] == ' '):
continue
if(expression[i]>='0' and expression[i] <= '9'):
charNumber = [] #for storing number
while(i<len(expression) and expression[i]>='0' and expression[i] <= '9'):
charNumber.append(expression[i])
i+=1
valueStack.append(int(''.join(charNumber)))
elif (expression[i]=='('):
opStack.append(expression[i])
elif (expression[i]==')'):
while(opStack[-1]!='('):
valueStack.append(self.applyOperation(opStack.pop(),valueStack.pop(),valueStack.pop()))
opStack.pop()
elif(expression[i]=='+'or expression[i]=='-'or expression[i]=='*'or expression[i]=='/'):
while( (len(opStack)!=0) and ( self.opPrecedence(expression[i],opStack[-1]) ) ):
valueStack.append(self.applyOperation(opStack.pop(),valueStack.pop(),valueStack.pop()))
opStack.append(expression[i])
i = i + 1
while(len(opStack)!=0):
valueStack.append(self.applyOperation(opStack.pop(),valueStack.pop(),valueStack.pop()))
return valueStack.pop()
def applyOperation(self,op,a,b):
if op=='+':
return a+b
elif op=='-':
return a-b
elif op=='*':
return a*b
elif op=='/':
return a/b
else:
return 0
def opPrecedence(self,op1,op2):
if (op2 == '(' or op2 == ')'):
return False
if ((op1 == '*' or op1 == '/') and (op2 == '+' or op2 == '-')):
return False
else:
return True
a = evaluateString()
print(a.evalString("(5+7)"))
I am able to get the right numbers in the valueStack. However, there seems to be problem in the last two elseif. Can someone point me in the right direction?

I have done some fixes and it works for some operations. But I haven't tested it for all cases. Also, operations are only integers, no floats (e.g. check last output below).
class evaluateString:
def evalString(self,expression):
valueStack = []
opStack = []
i=0
while(i<len(expression)):
if(expression[i] == ' '):
continue
if(expression[i]>='0' and expression[i] <= '9'):
charNumber = [] #for storing number
j = i
while(j<len(expression) and expression[j]>='0' and expression[j] <= '9'):
charNumber.append(expression[j])
j += 1
i = (j-1)
valueStack.append(int(''.join(charNumber)))
elif (expression[i]=='('):
opStack.append(expression[i])
elif (expression[i]==')'):
while(opStack[-1]!='('):
valueStack.append(self.applyOperation(opStack.pop(),valueStack.pop(),valueStack.pop()))
opStack.pop()
elif(expression[i]=='+'or expression[i]=='-'or expression[i]=='*'or expression[i]=='/'):
while( (len(opStack)!=0) and ( self.opPrecedence(expression[i],opStack[-1]) ) ):
valueStack.append(self.applyOperation(opStack.pop(),valueStack.pop(),valueStack.pop()))
opStack.append(expression[i])
i = i + 1
while(len(opStack)!=0):
valueStack.append(self.applyOperation(opStack.pop(),valueStack.pop(),valueStack.pop()))
return valueStack.pop()
def applyOperation(self,op,a,b):
if op=='+':
return a+b
elif op=='-':
return b-a
elif op=='*':
return a*b
elif op=='/':
return b/a
else:
return 0
def opPrecedence(self,op1,op2):
if (op2 == '(' or op2 == ')'):
return False
if ((op1 == '*' or op1 == '/') and (op2 == '+' or op2 == '-')):
return False
else:
return True
a = evaluateString()
print(a.evalString("8*12")) #prints 96
print(a.evalString("(122-434)")) #prints -312
print(a.evalString("(232+12)/2")) #print 122
print(a.evalString("232/12+2")) #prints 21

In python eval() will evaluate infix expressions
print(eval("(5+7)/2"))
it will print the evaluated infix expression value as 6.

Python invalid literal for int() with base 10: '+'

So first, the assignment given was to make the function "compute" solves it and returns the value. If the given sting is not a valid equation, returns "None"
def compute(side):
val=int(side[0])
lastop=0
for i in range(1,len(side)):
if side[i].isdigit():
if lastop=='+':
val+=int(side[i])
elif lastop=='-':
val-=int(side[i])
elif lastop=='x':
val*=int(side[i])
elif lastop=='/':
val/=int(side[i])
else:
lastop=side[i]
return val
So at this point the value would be returned. But if the function is run for ("22-11x4"), it gives 0 not 44. I've turned them into integers and why would they still give me the wrong value?
def evaluate():
val=int(side[0])
lastop=0
for i in range(1,len(side)):
if side[i].true():
print('Congrats')
elif side[i].false():
print('Try again')
And when it gets to this evaluate function, it gives the error of "invalid literal for int() with base 10: '+'" and I am not too sure what this means and how to solve.
def solve():
pass
This function "solve" I was going to get to later after I have fixed the problems in the functions before.
def main():
print("22-11x4 =", compute("22-11x4"),"(expect 44)")
print("+11x4 =", compute("+11x4"),"(expect None)")
print("22-11x4 ?= 7x5+9", evaluate("22-11x4=7x5+9"),"(expect True)")
print("solving 288/24x6=18x13x8 :", solve("288/24x6=18x13x8"))
main()

The compute function you wrote does operation by digit, also you forgot to do one, last operation after for was finished:
2 - 1 - 1 x 4 = 0
this why you got this return from the function
Here is the correct function
def compute(side):
val = 0
buf = 0
lastop = '+'
for i in range(0, len(side)):
if side[i].isdigit():
buf = buf * 10 + int(side[i])
else:
val = op(val, buf, lastop)
lastop = side[i]
buf = 0
val = op(val, buf, lastop)
return val
def op(val, buf, lastop):
if lastop == '+':
val += buf
elif lastop == '-':
val -= buf
elif lastop == 'x':
val *= buf
elif lastop == '/':
val /= buf
return val
Also in your testing string:
print("+11x4 =", compute("+11x4"),"(expect None)")
why do you expect None? it should be 44?
PS. this function could be improved yet, but I had no time to do it. anyway it works.

nesting a string and printing deepest nest

What I have is a string that looks for example like this {[]}{([])}()and I loop it through to find if it is an open or a close bracket and append it to the list.
What I wanna do now is to find which is the deepest nasting and print it out. So in this example i would print the middle {([ but I am confused how to do it. I can append the beginning of the open brackets and then reset it but how to compare couple of them though, and printing the biggest one
my code:
def is_nested(str):
stack = []
deepest =[]
index=1
open=0
for c in str:
if c == "{" or c == "(" or c == "[" or c =="<":
stack.append(c) # this is push
deepest.append(c)
open +=1
index +=1
elif c == "}":
x = stack.pop()
index +=1
if x != "{":
index -=1
x2=parens(x)
return "False: expected %s at character index %d, but received } instead." %(x2,index)
elif c == "]":
x = stack.pop()
index +=1
if x != "[":
index -=1
x2=parens(x)
return "False: expected %s at character index %d, but received ] instead." %(x2,index)
elif c == ">":
x = stack.pop()
index +=1
if x != "<":
index -=1
x2=parens(x)
return "False: expected %s at character index %d, but received > instead." %(x2,index)
elif c == ")":
x = stack.pop()
index +=1
if x != "(":
index -=1
x2=parens(x)
return "False: expected %s at character index %d, but received ) instead." %(x2,index)
check(str)
return True
def check(str):
deepest =[]
for c in str:
if c == "{" or c == "(" or c == "[" or c =="<":
deepest.append(c)
print deepest
def parens(x):
if x == "<":
return ">"
elif x =="(":
return ")"
elif x == "{":
return "}"
elif x == "[":
return "]"
print is_nested("{[()}")
print is_nested("([)")
print is_nested("{[]({})}")
print is_nested("<()>")
print is_nested("{(<)}")

Not sure if I understand your wishes correctly, but this finds the longest sequence of successive opening brackets:
In [20]: import re
In [21]: s = '{[]}{([])}()'
In [22]: max(re.findall("[\(\[\{]+",s),key=len)
Out[22]: '{(['

You should iterate and update the current number of open brackets and keep the max value that you had while looping. You can put all open brackets on a string that you use as a stack and update max with this string if the lenght is bigger than curren length of max.
OPEN = "<[({"
CLOSED = ">])}"
def is_nested(str):
stack = []
deepest =[]
for c in str:
if c in OPEN:
stack.append(c)
if len(stack)>len(deepest):
deepest.append(c)
elif c in CLOSED:
x = stack.pop()
if OPEN.index(x) != CLOSED.index(c):
return "Error"
return ''.join(deepest)

Something like this?
def is_nested(nested_str):
opening = ['(', '{', '[', '<']
closing = {'(':')','{':'}', '[':']', '<':'>'}
braces = []
depth = 0
max_depth = 0
max_depth_index = None
for index, char in enumerate(nested_str):
if char in opening:
braces.append(char)
depth += 1
elif char == closing[braces[-1]]:
braces.pop()
depth -= 1
else:
raise ValueError("This is not a valid str")
if depth > max_depth:
max_depth = depth
max_depth_index = index
return max_depth, max_depth_index
This function takes a string with only braces, and tells you the deepest level of nesting, and the index of the opening brace of the first instance of that level of nesting. As a bonus, it will raise an error if the string is miss-formed.
>>> is_nested('{[][]}')
(2, 1)
>>> is_nested('{[][<()()>]()}[]()')
(4, 5)
>>> is_nested('{(})')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "nesting.py", line 16, in is_nested
ValueError: This is not a valid str
I've defined any character other than a brace to be an error, but this could easily be changed by modifying the else condition.

def is_nested(in_str):
stack = []
deepest = tuple()
open_b = ('(','[','{','<')
close_b = (')',']','}','>')
for i in in_str:
print stack, i
if i in open_b:
stack.append(i)
if len(stack) > len(deepest):
deepest = tuple(stack)
else:
p = stack.pop()
print p
print open_b.index(p)
print close_b[open_b.index(p)]
if i != close_b[open_b.index(p)]:
raise Exception('Wrongly nested')
if len(stack) > 0:
raise Exception('Wrongly nested')
return deepest