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  1. Balanced Brackets
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Balanced Brackets

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250 Discussions

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  • g_cdcomp
     + 0 comments
    //JS Solution

function isBalanced(s) {
    let brackDict = [
        ["(",")"],
        ["[","]"],
        ["{","}"],
    ];
    
    
    let stack = [];
    
    for (let i = 0; i < s.length; i++) {
       let match = brackDict.find(item => item[0] == s[i]);
       if (match) {
            stack.push(match);
       } else {
            let last = stack.pop();
            if (last == undefined || last[1] != s[i]) {
              return 'NO';
            }
        }
    
    }  
    
    if (stack.length != 0) {
        return 'NO';
    }
    
    return 'YES';
}

  • evanmico222
     + 0 comments

    js answer using a map for checking correlation:

    function isBalanced(s) {
    // Write your code here
    let openStack = new Array();
    let idx = 0;
    let valid = true;
    let correlationMap = new Map([
        [')', '('],
        [']', '['],
        ['}', '{'],
    ]);
    while(valid && idx<s.length){
        if(correlationMap.has(s[idx]) && (correlationMap.get(s[idx]) != openStack.pop())){
            valid = false;
        }else if(!correlationMap.has(s[idx])){
            openStack.push(s[idx]);
        }
        idx++;
    }
    if(openStack.length > 0) valid = false;
    return (valid)?('YES'):('NO');
}

  • 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"