is_matched.py

#!/bin/python
""" hackerrank coding challenge to determine if a string of brackets is balanced or not """


def is_opposite(char1, char2):
    # if char1 is the opposite kind of bracket as char2 then return True
    # but this only works if char1 is of variety "LEFT bracket"
    # this means that order counts
    if char1 == "{" and char2 == "}":
        return True
    elif char1 == "[" and char2 == "]":
        return True
    elif char1 == "(" and char2 == ")":
        return True
    else:
        return False

def is_matched(expression):
    stack_string = []
    expression = [ x for x in expression ]
    # make the input string, expression, into a list object
    while len(expression) != 0:
        current_char = expression.pop(0)
        # above code pops one character from the left side of the input
        if current_char in '{[(':
            stack_string.append(current_char)
        elif current_char in ')}]' and len(stack_string) == 0:
            return False
        elif current_char in ')}]':
            if is_opposite(stack_string[-1],current_char):
                stack_string.pop()
        else:
            return False
    if len(stack_string) == 0:
        return True
    else:
        return False

t = int(raw_input().strip())
for a0 in xrange(t):
    expression = raw_input().strip()
    if is_matched(expression) == True:
        print "YES"
    else:
        print "NO"