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  1. Balanced Brackets
  2. Discussions

Balanced Brackets

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40 Discussions

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  • vadym_usatiuk
     + 0 comments

    Java O(n)

    public static String isBalanced(String s) {
        Stack<Character> stack = new Stack<>();

        for (char c : s.toCharArray()) {
            if (c == '(' || c == '{' || c == '[') {
                stack.push(c);
            } else {
                if (stack.isEmpty()) {
                    return "NO";
                }
                char top = stack.pop(); 
                if ((c == ')' && top != '(') || (c == '}' && top != '{') || (c == ']' && top != '[')) {
                    return "NO";
                }
            }
        }
        return stack.isEmpty() ? "YES" : "NO";
    }

  • vitorio_tfm
     + 0 comments

    Java 15

    var squareOpen = false;
var curlyOpen = false;
var parenthesisOpen = false;

Queue<Character> brackets = new LinkedList<>();
Stack<Character> expectedClosures = new Stack<>();
for (var c : s.toCharArray()) {
	brackets.add(c);
}

while (!brackets.isEmpty()) {
	var ele = brackets.poll();
	try {
		switch (ele) {
			case '[': squareOpen = true; expectedClosures.add(']'); break;
			case '(': parenthesisOpen = true; expectedClosures.add(')'); break;
			case '{': curlyOpen = true; expectedClosures.add('}'); break;
			case ']': squareOpen = false; if (expectedClosures.pop() != ele) return "NO"; break;
			case ')': parenthesisOpen = false; if (expectedClosures.pop() != ele) return "NO"; break;
			case '}': curlyOpen = false; if (expectedClosures.pop() != ele) return "NO";
		}
	} catch (EmptyStackException ex) {
		return "NO";
	}
}
return !squareOpen && !parenthesisOpen && !curlyOpen ? "YES" : "NO";

  • pnsemonov
     + 0 comments

    JS

    function isBalanced(string) {
    const pairs = new Map([[")", "("], ["}", "{"], ["]", "["]]);
    const brackets = [];
    let balanced = true;
    for (let bracket of string) {
        if (pairs.has(bracket)) {
            if (brackets.pop() != pairs.get(bracket)) {
                balanced = false;
                break;
            }
        }
        else {
            brackets.push(bracket);
        }
    }
    return (balanced && brackets.length == 0) ? "YES" : "NO";
}

  • guysneh
     + 0 comments

    C#

    char last = ' ';
  for (int i = 0;i<s.Length;i++) 
  {
    char c = s[i];
    switch (c) 
    {
      case '{':
        break;
      case '[':
        break;
      case '(':
        break;
      case ')':
        if (last=='(') 
        {
          s = s.Remove(i - 1, 2);
          i -= 2;
        }
        else 
        {
          return "NO";
        }
        break;
      case ']':
        if (last == '[')
        {
          s = s.Remove(i - 1, 2);
          i -= 2;
        }
        else
        {
          return "NO";
        }
        break;
      case '}':
        if (last == '{')
        {
          s = s.Remove(i - 1, 2);
          i -= 2;
        }
        else
        {
          return "NO";
        }
        break;
    }
      if (i>-1) 
    {
      last = s[i];
    }     
    else 
    {
      last = ' ';
    }
  }

  if (s.Length==0) 
  {
    return "YES";
  }
  else 
  {
    return "NO";
  }

  • trinhhuynhthinh1
     + 0 comments

    Python

    def getKey(c):
    if(c == '{'): return 1
    if(c == '}'): return 2
    if(c == '('): return 3
    if(c == ')'): return 4
    if(c == '['): return 5
    if(c == ']'): return 6

    
def isBalanced(s):
    # Write your code here
    arrCharacter  = list(s)
    stack = []
    for x in arrCharacter:
        if(len(stack)==0 or getKey(x) != 1+getKey(stack[-1])):
            stack.append(x)
        else:
            stack.pop()
    print(stack)
    return "YES" if len(stack)==0 else "NO"