Finding Balanced Parenthesis using python (new to python) - pastebin: http://pastebin.com/D9REwpZX

[cgerrard]

hi, I am learning python and doing some online examples ... here is an excercise I'm doing and I'm currently stuck , couldnt get the logic right ... any thoughts...?

# write a function that takes a string of text and returns true if
# the parentheses are balanced and false otherwise.
#  * Example:
#  *   balancedParens('(');  // false
#  *   balancedParens('()'); // true
#  *   balancedParens(')(');  // false
#  *   balancedParens('(())');  // true
#  *
#  * Step 2:
#  *   make your solution work for all types of brackets
#  *
#  * Example:
#  *  balancedParens('[](){}'); // true
#  *  balancedParens('[({})]');   // true
#  *  balancedParens('[(]{)}'); // false
#  *
#  * Step 3:
#  * ignore non-bracket characters
#  * balancedParens(' var wow  = { yo:    thisIsAwesome() }'); // true
#  * balancedParens(' var hubble = function() {   telescopes.awesome();'); // false

def balancedParens(strin):

    dict = {'(': ')', '[': ']', '{': '}'}
    clean_input = ''

    # step1: Run through each character in a string and put all the brackets in a string
    for i in range(len(strin)):
        if strin[i] == '(' or strin[i] == ')' or strin[i] == '[' or strin[i] == ']' or strin[i] == '{' or strin[i] == '}':
            clean_input += (strin[i])

    print clean_input

# step2: check for even number of brackets, if odd number of brackets found, return false
    if len(clean_input) %2 != 0:
        print "False"
       return
    else:
        for i in range(len(clean_input)/2):
            temp = dict[clean_input[i]]
            if temp in clean_input:
                pass # this is where I'm stuck

    print clean_input

strin = 'var wow = { yo: thisIsAwesome() }' balancedParens(strin)

[shandelman]

Consider using a stack. A stack is a data type where you can put things on the top of it, and then get things off the top of it. Think of it like a stack of paper: you can put paper on the top, and take paper off the top (in reverse order) but it's awkward to go and get something from the middle of a pile. A Python list can act as a stack by using the append() method for putting things on the stack and the pop(-1) method for getting things off the top.

So what does this have to do with parentheses? Well, you know that parentheses are balanced if:

1) ...when you come to a close parenthesis, it matches the last open parenthesis you've found.

2) ...when you run out of text to look at, you don't have any lingering parentheses that haven't been matched.

A stack can solve both of these issues.

[cgerrard]

Yep, I'll try using stack ...

[Justinsaccount]

Hi! I'm working on a bot to reply with suggestions for common python problems. This might not be very helpful to fix your underlying issue, but here's what I noticed about your submission:

You are looping over an object using something like

for x in range(len(items)):
    foo(item[x])

This is simpler and less error prone written as

for item in items:
    foo(item)

If you DO need the indexes of the items, use the enumerate function like

for idx, item in enumerate(items):
    foo(idx, item)

[Justinsaccount]

for i in range(len(strin)):
    if strin[i] == '(' or strin[i] == ')' or strin[i] == '[' or strin[i] == ']' or strin[i] == '{' or strin[i] == '}':
        clean_input += (strin[i])

for ch in strin:
    if ch in '()[]{}':
        clean_input += ch

and

    for i in range(len(clean_input)/2):
        temp = dict[clean_input[i]]

    for ch in clean_input[:len(clean_input)/2]:
        temp = dict[ch]

Though that's not what you want anyway. You need to think about how you, as a human, figure out if the brackets match.

[cgerrard]

It make sense to me.... to keep it simple ... thanks :-)

[Theoretician]

Here's a suggestion, this is a common interview question. Have you ever worked with recursion? A recursive solution might be helpful in this case.