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

Balanced Brackets

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1635 Discussions

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  • ale_220614
     + 0 comments

    JAVA 8

    public static String isBalanced(String s) { String response = "YES"; String[] values = s.split(""); String listExpected = ""; String passValues = "({[";

    for(int i = 0; i < values.length; i ++){
    String value = values[i];

    if(!passValues.contains(value)){
        String[] list = listExpected.split("");
        if(list[0].equals(value)){
            listExpected = listExpected.replaceFirst("\\"+value, "");
        } else {
            response = "NO";
            break;
        }
    }

    if(value.equals("(")){
        listExpected = ")" + listExpected;
    }

    if(value.equals("{")){
        listExpected = "}" + listExpected;
    }

    if(value.equals("[")){
        listExpected = "]" + listExpected;
    }
}

return response;
}

  • bonnemai
     + 0 comments
    def isBalanced(s):
    _map={')': '(', ']':'[', '}':'{'}
    q=[] 
    for character in s:
        if character in '([{': 
            q.append(character)
        elif character in ')]}': 
            if not q: 
                return 'NO'
            last=q.pop()
            if last!=_map[character]: 
                return 'NO'
    return 'YES' if not q else 'NO'

  • xavi_aclm
     + 0 comments

    use a lifo queue and scan left to right

  • gauravvgaikwad07
     + 0 comments

    What's the issue?

    include

    using namespace std; void printstack(stack s){ while(!s.empty()){ cout< st; bool t=true; for (auto ch:s){ if ((ch=='(')||(ch=='[') || (ch=='{')){ st.push(ch); //printstack(st); }else{ if ((ch==')') && (st.top()=='(')){ st.pop(); }else if((ch==']') &&(st.top()=='[')){ st.pop(); }else if((ch=='}') && (st.top()=='{')){ st.pop(); }else{ return "NO"; } } }

    if (st.empty()){
    return "YES";
}else{
    return "NO";
}

    } int main(){ int t; cin>>t; while(t--){ string s; cin>>s; cout<

  • f97gp1
     + 0 comments

    I try a recursive solution just for fun (won't pass the test cases)

        # rest of the code ...
    # the methods are naive methods and self explained.
    # `stack` variable could be considered as a global variable.
    def isBalanced(self, s: str, stack: deque) -> bool:
        c: str = s[0]
        
        if self.isOpenBracket(c):
            stack.append(c)
        else:
            stack_top: str = stack.pop()
            if not self.areBothBracketsBalanced(stack_top, c):
                print('\tnot-match: ', stack_top, c)

                return False
            
            if len(s) == 1 and len(stack) == 0:
                print('\tfinal-case OK')
                
                return True
        
        # New solution.
        new_sol: str = s[1:]
        
        if len(new_sol) > 0:
            return self.isBalanced(new_sol, stack)