Balanced Brackets Discussions |

Sort 184 Discussions, By:

1 day ago+ 0 comments

There's probably a more efficient solution than this, but this worked for me (in Python).

x = s if len(x) % 2 != 0: return 'NO'

a = '()'
b = '[]'
c = '{}'

while a in x or b in x or c in x:
    x = x.replace(a, '')
    x = x.replace(b, '')
    x = x.replace(c, '')

if len(x) != 0:
    return 'NO'

return 'YES'

2 days ago+ 0 comments

Try to make it clear with python:

def isBalanced(s):
    if s[0] in ['}', ']', ')']:
        return 'NO'
    dic = {'}':'{', ']':'[', ')':'('}
    stack = []
    
    for i in s:
        if i in ['{', '[', '(']:
            stack.append(i)
        else:
            if not stack:
                return 'NO'
            value = stack.pop() 
            if value != dic[i]:
                return 'NO'
    if stack:
        return 'NO'
    return 'YES'

2 days ago+ 0 comments

Yet another python. I was sure I'll find here much more sophisticated code when finally complete the task by myself and I was right :D Anyway, this is my solution:

def isBalanced(s):
    # Write your code here
    print(s)
    tempTab = []
    tempValue = ""
    if (len(s) % 2) != 0:  # check if every bracket has a pair
        return "NO"
    if (s[0] == '(' or s[0] == '{' or s[0] == '['):  # check first element
        tempTab.append(s[0])
    else:
        return "NO"
    for x in range(1, len(s)):  # check the rest of elements
        if (s[x] == '(' or s[x] == '{' or s[x] == '['):  # if opening bracket, add it to the list
            tempTab.append(s[x])
        else:  # otherwise, check if the last opening bracket fits to the current closing bracket
            try:
                tempValue = tempTab.pop()  # remove the last opening
            except IndexError:
                return "NO"
            if s[x] == ')':
                if (tempValue == '('):
                    pass
                else:
                    return 'NO'
            elif s[x] == '}':
                if (tempValue == '{'):
                    pass
                else:
                    return 'NO'
            elif s[x] == ']':
                if (tempValue == '['):
                    pass
                else:
                    return 'NO'
    if len(tempTab) == 0:  # finally, check if all the opening brackets have been closed
        return 'YES'
    else:
        return 'NO'

3 days ago+ 0 comments

My Python solution using a stack:

def isBalanced(s: str) -> str:
     
    def match_bracket(c, stack):
        c_match = stack.get()
        if c=='}' and c_match=='{':
            return stack
        elif c==']' and c_match=='[':
            return stack
        elif c==')' and c_match=='(':
            return stack
        else:
            stack.put(c_match)
            stack.put(c)
            return stack
            
    s_stack = LifoQueue()
    for c in s:
        if s_stack.empty():
            s_stack.put(c)
            continue
        else: 
            s_stack = match_bracket(c, s_stack)
            
    return 'YES' if s_stack.empty() else 'NO'

2 weeks ago+ 0 comments

function isBalanced(s) {
    // Write your code here
    let brackets = s.split('')
    let closedBrackets = []
    let currentBraket
    let closedBracket

    while(brackets.length > 0){
      currentBraket = brackets.pop()
      // console.log(brackets,currentBraket)
      
      if(currentBraket === ')' | currentBraket === '}'| currentBraket === ']'){
        closedBrackets.push(currentBraket)
      }else{
        if(closedBrackets.length == 0){
          return 'NO'
        }else{
          closedBracket = closedBrackets.pop()
          if(currentBraket == '(' && closedBracket !== ')') return 'NO'
          if(currentBraket == '{' && closedBracket !== '}') return 'NO'
          if(currentBraket == '[' && closedBracket !== ']') return 'NO'
        }
      }
    }
    if(closedBrackets.length > 0) return 'NO'
    return 'YES'
}

Load more conversations