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

Balanced Brackets

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  • ashokprajpati
     + 0 comments

    JAVASCRIPT - AP

       let stack = [];
   let ans = "YES";
   
   let check = (b) => {
        if(stack[stack.length-1]==b){
            stack.pop();
            return true
        }
        ans="NO";
        return false;
   };
   
   for(let el of s){
       if(el==="{" || el==="[" || el ==="(") stack.push(el);
       else if(el==="}"){ if(!check("{")) return ans} 
       else if(el==="]"){ if(!check("[")) return ans} 
       else if(el===")"){ if(!check("(")) return ans}
   }
   
    if(stack.length) return "NO";
    return ans;

  • Rs_poonia
     + 0 comments

    JavaScript simple, easiest and best solution

    function isBalanced(s) { // Write your code here let map = { "(": ")", "{":"}", "[": "]" } const res = [] for(let i = 0; i

  • tarunsk56
     + 0 comments

    def isBalanced(s): # Write your code here stk = [] for i in s: if i == '(' or i == '{' or i == '[': stk.append(i) else: if len(stk) == 0: return "NO" elif i == ')' and stk[-1] == '(': stk.pop() elif i == '}' and stk[-1] == '{': stk.pop() elif i == ']' and stk[-1] == '[': stk.pop() else: return "NO"

    if len(stk)>0:
    return "NO"
return "YES"
            return "NO"

if len(stk)>0:
    return "NO"
return "YES"

  • khoai345678
     + 0 comments

    python implementation

    def isBalanced(s):
    check = {")":"(" , "]":"[" , "}":"{" }
    stack = []
    for c in s:
        if c in check.values():
            stack.append(c)
        elif stack and check[c] == stack[-1]:
            stack.pop()
        else:
            return 'NO'
    if stack == []:
        return 'YES'
    else:
        return 'NO'

  • shivang_sharma
     + 0 comments

    C++ Solution

    string isBalanced(string s) {
    stack<char> brackets{};
    map<char, char> pairs = {
        {')', '('},
        {'}', '{'},
        {']', '['},
    };
    
    for(auto& ch: s){
        if(pairs.find(ch) == pairs.end()){
            brackets.push(ch);
        }else{
            if(!brackets.empty() && brackets.top() == pairs[ch]){
                brackets.pop();
            }else{
                return "NO";
            }
        }
    }
    
    if(brackets.size())
        return "NO";
    else
        return "YES";
}
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