CareerCup

using a stack, if we see opening P, B or C, push, else if we see closing sign, do a pop, compare the two sign, if they are from same P, B or C, Correct.

keep a counter for Parentheses, Bracket, and Curly.
let p = 0;
let b = 0;
let c = 0;

scanning from left to right on our sequence, if we see opening p or b or c, we increment by 1 the appropriate variable and decrement by 1 if we see a closing P or B or C.

From the above examples are not enough.

We check if the string is correct or not every time we see

1) A CLOSING sign. There are a few cases here.

2) an end of string character. p|c|b should all be zeros.

// psuedo code
bool Match(char token)
{
   char c;
   switch( c=ReadToken() )
   {
      case '{':
      case '[':
      case '(':
            if(Match(c))
               return Match(token);
            return false;

      case '}':
           return (token=='{');
      case ']':
           return (token=='[');
      case ')':
           return (token=='(');

      case END:
           return false;           

      default:
           return Match(token);


   }
}

U can use a stack as luv said....
Push all the opening and when you see a closing then pop from the stack ....then you should see the corresponding else it is unmatched.. do it till the stack is empty.

int check(char* s){
	if(s==NULL)
		return 1;
	Stack s=new Stack();
	s.push(s[0]);
	int i=1;
	while(!s.isempty()){
		if(s[i] =='[' || '{' || '(')
			s.push(s[i];
		else{
			char t=s.pop();
			if(!(s[i]==']' && t=='[') && (s[i]=='}' && t=='{') && (s[i]==')' && t=='('))
				return 0;
		}

	}
	return 1;
}

Sorry it is
if(!(s[i]==']' && t=='[') || (s[i]=='}' || t=='{') || (s[i]==')' && t=='('))
return 0;

vamsi's soln is correct.