Balanced brackets

Have you observed something? When ever you are writing code using any IDE, if you write mismatched brackets, immediately an error is shown by IDE.

So how does IDE  know if an expression is having balanced brackets? For that, the expression must have equal number of opening and closing brackets of matching types and also they must be in correct order.

Let us look at some examples

(a+b)*c+d*{e+(f*g)}   - balanced

(p+q*[r+u)] - unbalanced

(p+q+r+s)) - unbalanced

(m+n*[p+q]+{g+h}) - balanced


So we do we write a program to check if an expression is having balanced brackets? We do need to make use of stack to store the brackets.

The algorithm is as follows

1. Scan a character - ch from the expression
2. If the character is opening bracket, push it to stack
3. If the character is closing bracket
1. pop a character from stack
2. If popped opening bracket and ch are not of same type ( ( and ) or [ and ] ) stop the function and return false
4. Repeat steps 2 and 3 till all characters are scanned.
5. Once the loop is completed, if the stack is empty return true else return false

Once we have this algorithm, code will be quite simple.

int is_opening_bracket(char ch)
{
    switch(ch)
    {
	case '(':
	case '[':
	case '{':
	 
		return 1;
        default : return 0;
    }
}

int is_closing_bracket(char ch)
{
    switch(ch)
    {
	case ')':
	case ']':
	case '}':
	 
		return 1;
        default : return 0;
    }
} 
 
int is_matching_bracket(struct node **top,char ch)
{
    char ch2 = pop(top);
    if(ch2!=ERRORCHAR)
    {
    switch(ch)
    {
	case ')': if(ch2!='(') return 0;
                  break;
	case ']': if(ch2!='[') return 0;
                    break;
	case '}': if(ch2!='{') return 0;
                    break;
	 
		return 1;
        default : return 0;
    }
   }else
    return 0;
} 
int is_balanced(char *expression)
{
     char ch;
      
     struct node *top = NULL;
     while(ch =*expression++)
     {
	if(is_opening_bracket(ch))
        {
            top = push(ch,top);
        }
        else if(is_closing_bracket(ch))
	{
	  if(!is_matching_bracket(&top,ch))
          {
            return 0;           
          }
	}
     }
     return is_empty(top);
}
  

I have used 3 helper function to find out if the character is opening bracket, is a closing bracket and if the character matches with the bracket popped from stack. The stack in this case a character stack.

And here you have the driver program with stack implemented using linked list.

#include<stdio.h>
#include<string.h>
#include<stdlib.h> 
#define ERRORCHAR 65
struct node
{
    char ch;
    struct node *next;
};
struct node *createnode(char ch)
{
    struct node *newnode = (struct node*) malloc(sizeof(struct node));
    newnode->ch = ch;
    newnode->next = NULL;
}

int is_empty(struct node *top)
{
    return top==NULL;
}
struct node * push(char ch,struct node *top)
{
   struct node *newnode = createnode(ch);
   newnode->next = top;
   top = newnode;
   return top;
}

char   pop(struct node ** top)
{
   char ch;
   if(*top==NULL)
	return ERRORCHAR;
   struct node * temp = *top;
   *top =(*top)->next;
   ch = temp->ch;
   
   free(temp);
   return ch;
}
   
int main()
{
   char expression[50];
   double ans;
   printf("Enter an expression :");
   scanf("%s",expression); 
   if(is_balanced(expression))
   {
        printf("The expression has balanced brackets");
   }
   else
   {
        printf("The expression has unbalanced brackets");
   }
   return 0;
}