Balanced Brackets| Stack
TASK(easy)
A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and ().A matching pair of brackets is not balanced if the set of brackets it encloses are not matched. For example, {[(])} is not balanced because the contents in between { and } are not balanced. The pair of square brackets encloses a single, unbalanced opening bracket, (, and the pair of parentheses encloses a single, unbalanced closing square bracket, ].
By this logic, we say a sequence of brackets is balanced if the following conditions are met:
- It contains no unmatched brackets.
- The subset of brackets enclosed within the confines of a matched pair of brackets is also a matched pair of brackets.
Given n strings of brackets, determine whether each sequence of brackets is balanced. If a string is balanced, return YES. Otherwise, return NO.
Function Description
Complete the function isBalanced in the editor below.
isBalanced has the following parameter(s):
- string s: a string of brackets
Returns
- string: either YES or NO
Input Format
The first line contains a single integer n, the number of strings.
Each of the next lines n contains a single string s, a sequence of brackets.
Constraints
All characters in the sequences ∈ { {, }, (, ), [, ] }.
Output Format
For each string, return YES or NO.
Sample Input
STDIN Function
----- --------
3 n = 3
{[()]} first s = '{[()]}'
{[(])} second s = '{[(])}'
{{[[(())]]}} third s ='{{[[(())]]}}'
Sample Output
YES
NO
YES
SOLUTION 1
def isBalanced(s):
refi=['{', '[', '(', ')', ']', '}']
refe=['}', ']', ')']
z=list()
if len(s) == 0:
return 'YES'
if len(s) % 2 == 1:
return 'NO'
if refi.index(s[0])>2 or refi.index(s[-1])<=2:
return 'NO'
for i in s:
if refi.index(i)<3:
z.append(refi.index(i))
else:
if z and z[-1]-refe.index(i)==0:
z.pop()
else:
return 'NO'
if not z:
return 'YES'
