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  1. Prepare
  2. Data Structures
  3. Stacks
  4. Balanced Brackets
  5. Discussions

Balanced Brackets

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  • eren_akyurek_cse
     + 0 comments

    C++14

    bool isMatching(char a, char b){
    if (a=='{' && b== '}')
    return 1;
    else if (a=='(' && b== ')')
    return 1;
    else if (a=='[' && b== ']')
    return 1;
    else return 0;
}



string isBalanced(string s) {
    stack<char> st;
    
    for (int i = 0; i < s.length(); i++){
        if (s[i] == '(' || s[i] == '[' || s[i] == '{' ){
            st.push(s[i]);
        }
        else if(st.empty() || !isMatching(st.top(), s[i])){
            return "NO";
        }
        else{
            st.pop();
        }
    }return st.empty() ? "YES" : "NO";

}

  • cherrymeteor
     + 0 comments

    JS

    function isBalanced(s) {
  let stack = [], brackets = { "}": "{", "]": "[", ")": "(" }
  for (let str of s.split("")) {
    if (Object.values(brackets).includes(str)) stack.push(str)
    else if (stack[stack.length - 1] == brackets[str]) {
      stack.pop()
    } else {
      return "NO"
    }
  }
  return stack.length == 0 ? "YES" : "NO"
}

  • abdulquadrioyek1
     + 0 comments
    #python3
def isBalanced(s):
    stack = []
    closeToOpen = {
        '}': '{',
        ')': '(',
        ']': '['
    }
    for c in s:
        if c in closeToOpen:
            if stack and stack[-1] == closeToOpen[c]:
                stack.pop()
            else:
                return 'NO'
        else:
            stack.append(c)
    return 'YES' if not stack else 'NO'

  • bosymm
     + 0 comments

    JAVA

    public static String isBalanced(String s) {
    // Write your code here
    Stack<Character> stack = new Stack<>();

    for(int i=0; i<s.length(); i++) {
        if(s.charAt(i) == '(') {
            stack.push(')');
        } else if(s.charAt(i) == '{'){
            stack.push('}');
        } else if (s.charAt(i) == '[') {
            stack.push(']');
        } else if( stack.isEmpty() || s.charAt(i) != stack.pop() ) {
            return "NO";
        }
    }

    return stack.isEmpty() ? "YES" : "NO";
}

  • trietkd1475369
     + 0 comments
    def isBalanced(s):
    stack=[]
    for char in s:
        if char=='{' or char=='(' or char=='[':
            stack.append(char)
        if char==']' and (len(stack)==0 or (len(stack)!=0 and stack[-1]!='[')):
            return "NO"
        if char=='}' and (len(stack)==0 or (len(stack)!=0 and stack[-1]!='{')):
            return "NO"
        if char==')' and (len(stack)==0 or (len(stack)!=0 and stack[-1]!='(')):
            return "NO"
        if char=='}' and len(stack)>0 and stack[-1]=='{':
            stack.pop(-1)
        if char==']' and len(stack)>0 and stack[-1]=='[':
            stack.pop(-1)
        if char==')' and len(stack)>0 and stack[-1]=='(':
            stack.pop(-1)
    if len(stack)==0:
        return "YES"
    return "NO"
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