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  1. Balanced Brackets
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Balanced Brackets

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252 Discussions

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  • p_c_c_o_d_e
     + 0 comments

    How is this problem intermediate yet there are absolute insane problems that take me 30-45 minutes to even understand what they're asking under "easy"?

    This is the easiest question I've ever answered and an ACTUAL REAL WORLD PROBLEM UNLIKE 95% OF THE BS ON THIS SITE.

  • nandun5
     + 0 comments
        public static string isBalanced(string s)
    {
        Stack<char> brackets = new Stack<char>();
        
        foreach(char bracket in s){
            if(brackets.TryPeek(out char last) && last == Opposite(bracket)) brackets.Pop();
            else brackets.Push(bracket);
        }
        
        if(brackets.Any()) return "NO";
        
        return "YES";
    }
    
    public static char Opposite (char bracket){
        switch(bracket){
            case '}' : return '{';
            case ')' : return '(';
            case ']' : return '[';            
        }
        return default;
        
    }

  • 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";

    }