We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Balanced Brackets
  2. Discussions

Balanced Brackets

Sort by

recency

|

248 Discussions

|

  • basilleno
     + 0 comments
    

    stack<char> Brackets;
for(char c : s)
{
    if(c == '(' || c == '[' || c == '{')
        Brackets.push(c);
    else 
    {

        if(Brackets.empty()) return "NO";
        char top = Brackets.top();
        if((c == ')' && top != '(')
        || (c == ']' && top != '[')
        || (c == '}' && top != '{'))
            return "NO";
        else 
            Brackets.pop();   
    }
}
        return Brackets.empty() ? "YES" : "NO";

    }

  • richardberich
     + 0 comments

    The thing to understand is that at any step, when a closing bracket is met, it must be a match for the last encountered opening bracket. Using a stack to track expected next closing bracket does the job .

  • uchihanigobi
     + 0 comments
    def isBalanced(s):
    # Write your code here
    open_bracket = "{[("
    enclose_bracket = "}])"
    pairs = ["{}", "[]", "()"]
    
    if len(s) % 2 != 0:
        return "NO"
    
    if s[0] in enclose_bracket or s[-1] in open_bracket:
        return "NO"
    
    stack = []
    for p in s:
        if p in open_bracket:
            stack.append(p)
        else:
            if len(stack) == 0:
                return "NO"
                
            pair = stack.pop() + p
            if pair not in pairs:
                return "NO"
    
    return "YES"

  • wasifk308
     + 0 comments
    string isBalanced(string s) {
stack<char> stk;


for (int i = 0 ; i<s.length(); i++) 
{
if(!stk.empty())
{
char tp  =stk.top();

if( (tp == '{' && s[i] == '}')||(tp == '(' && s[i] == ')') ||( tp == '[' && s[i] == ']'))
{
stk.pop();

}else {
stk.push(s[i]);
}
}else {
stk.push(s[i]);

}

}

  • anw_g01
     + 1 comment

    Python:

    def isBalanced(s):
    left, right = ["(", "{", "["], [")", "}", "]"]
    stack = []
    for char in s:
        # add opening brackets to the stack
        if char in left:
            stack.append(char)
        # if it's a right bracket:
        else:       
            # closing bracket appears without an opening bracket
            if not stack:
                return "NO"     
            # check the last opening bracket in the stack
            idx = right.index(char) 
            if stack.pop() == left[idx]:    # remove last bracket in-place
                continue
            return "NO"    # closing bracket didn't find an opening match
    return "NO" if stack else "YES"     # "YES" if stack is empty