Balanced Brackets Discussions |

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3 days ago+ 0 comments

Python 3 solution

def isBalanced(s):
    op = []
    for i in range(len(s)):
        if s[i] in '({[':
            op.append(s[i])
        elif s[i] in ')}]':
            try:
                if op[-1]+s[i] == '()' or op[-1]+s[i] == '{}' or op[-1]+s[i] == '[]':
                    op.pop(-1)
                else:
                    return 'NO'
            except:
                return 'NO'
    if len(op) != 0:
        return 'NO'
    else:
        return 'YES'

5 days ago+ 0 comments

JavaScript Solution

function isBalanced(s) {
    let stack=[], top=-1, map=new Map([[')','('],['}','{'],[']','[']])
    for(let i=0;i<s.length;i++){
        if(/[\)\}\]]/.test(s[i])){
            if(top>=0 && map.get(s[i])==stack[top] ) top--
            else return "NO"
        }
        else stack[++top]=s[i]
    }
    return top>=0?"NO":"YES"
}

2 weeks ago+ 0 comments

Java Eassy: import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
    tests:
    for(int a0 = 0; a0 < t; a0++){
        String s = in.next();
        Stack<Character> stack = new Stack<>();

        for(char c : s.toCharArray())
        {
            if(c == '(')
                stack.push(')');
            else if(c == '{')
                stack.push('}');
            else if(c == '[')
                stack.push(']');

            else{
                if( stack.isEmpty() || c != stack.peek()){
                    System.out.println("NO");
                    continue tests;    
                }
                else{
                    stack.pop();
                }
            }
        }
        if(stack.isEmpty())
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}

}

2 weeks ago+ 0 comments

    public static string isBalanced(string s)
    {
        Stack<char> left = new Stack<char>();
        int lastLeftIndex = 0;
        for(int i = 0; i < s.Length; i++)
        {
            if(s[i] == '(' || s[i] == '{' || s[i] == '[')
            {
                left.Push(s[i]);
            }
            else if(s[i] == ')' && (left.Count == 0 || left.Pop() != '('))
            {
                return "NO";
            }
            else if(s[i] == '}' && (left.Count == 0 ||left.Pop() != '{'))
            {
                return "NO";
            }
            else if(s[i] == ']' && (left.Count == 0 ||left.Pop() != '['))
            {
                return "NO";   
            }
        }
        if(left.Count > 0)
        {
            return "NO";
        }
        return "YES";
    }

3 weeks ago+ 0 comments

Python Recursive solution

def checkString(s): 
    # function to check if there's a match 
    original = s
    s = s.replace("{}","")
    s = s.replace("[]","")
    s = s.replace("()","")
    if s==original:
        return None 
    else: 
        return s
    
def isBalanced(s):
    # Write your code here
    while (len(s)!=0):
        s = checkString(s) 
        if s is None: 
            return "NO" 
        
    return "YES"

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