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  1. Balanced Brackets
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Balanced Brackets

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233 Discussions

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  • lynx3535
     + 0 comments

    C++ O(n)

    the idea:

    • each time you encounter an opening bracket, push it to the stack. each time you encounter a closing bracket, check if the opening bracket at the top of the stack matches the closing bracket, if so, pop the stack.
    • to prevent segmentation fault in case of encountring a closing bracket with an empty stack, simply return NO. this would for example here: }(){

    The code:

    string isBalanced(string s) {
    stack<char> stk;
    char top;
    for(char c : s){
        if(c == '(' || c == '[' || c == '{')
            stk.push(c);
						
        if(c == ')' || c == ']' || c == '}'){
            if(stk.empty()) return "NO";
            top = stk.top();
						
            if(c == ')' && top == '(' || c == ']' && top == '[' || c == '}' && top == '{')
                stk.pop();
								
            else
             return "NO";
        }
             
    }
    return stk.empty() ? "YES" : "NO";
}

  • dcfrostad
     + 1 comment

    Using a stack is unnecessary. We can simply match pairs of brackets using Regex and remove them from the string until the string is gone. If we ever don't find a match or if the string length is not an even number we can return 'NO'

    Here's a Javascript solution:

    function isBalanced(s) {
    do {
        const n = s.replace(/(\[\]|{}|\(\))/g, '')
        if (n.length === s.length || s.length % 2) {
            return 'NO'
        }
        s = n
    } while (s.length)
    return 'YES'
}

  • tjmfollosco
     + 0 comments

    Tried it again. I don't know why but I got better without coding for two days. Well C# at least. I was still making a website for the past 2 days. It's now 100%

    public static string isBalanced(string s)
    {
        int count = 0;
        Stack<char> myStack = new Stack<char>();
        var myDict = new Dictionary<char,char>();
        myDict.Add(')','(');
        myDict.Add('}','{');
        myDict.Add(']','[');
        
        for(int i = 0; i < s.Length; i++){
            if(count <= 0){
                if(myDict.ContainsKey(s[i])){
                    return "NO";
                    break;
                }
                myStack.Push(s[i]);
                count++;
            } else {
                if(myDict.ContainsKey(s[i])){
                    char temp = myStack.Pop();
                    if(temp != myDict.GetValueOrDefault(s[i])){
                        return "NO";
                        break;
                    }
                    count--;
                    
                } else {
                    myStack.Push(s[i]);
                    count++;
                }
            }
        }
        if(count > 0)
            return "NO";
        return "YES";
    }

  • tjmfollosco
     + 0 comments

    How the hell is this wrong?

    public static string isBalanced(string s)
    {
        bool isBalanced = true;
        int count = 0;
        Console.WriteLine(s);
        Stack myStack = new Stack();
        Dictionary<char,char> myDict = new Dictionary<char,char>();
        myDict.Add(')','(');
        myDict.Add('}','{');
        myDict.Add(']','[');
        for(int i = 0; i < s.Length; i++){
            if(myDict.ContainsKey(s[i]) && count > 0)
            {
                if(Convert.ToChar(myStack.Pop())!=myDict.GetValueOrDefault(s[i]))
                    isBalanced = false;
                count--;
            }
            else
            {
                 myStack.Push(s[i]);
                 count++;
            }
        }
        
        return (isBalanced) ? "YES" : "NO";
    }

  • gtai_quant
     + 0 comments
    def isBalanced(s):
    # Write your code here
    stack = []
    left_brackets = '([{'
    bracket_map = {')': '(', ']': '[', '}': '{'}
    for c in s:
        if c in left_brackets:
            stack.append(c)
        else:
            if stack and stack[-1] == bracket_map[c]:
                stack.pop()
            else:
                return 'NO'
    return 'NO' if stack else 'YES'