balanced brackets

 Problem:-

A bracket is considered to be any one of the following characters: (, ), {, }, [, or ].

Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {},and ().

A matching pair of brackets is not balanced if the set of brackets it encloses are not matched. For example, {[(])} is not balanced because the contents in between { and } are not balanced. The pair of square brackets encloses a single, unbalanced opening bracket, (, and the pair of parentheses encloses a single, unbalanced closing square bracket, ].

By this logic, we say a sequence of brackets is balanced if the following conditions are met:

It contains no unmatched brackets.
The subset of brackets enclosed within the confines of a matched pair of brackets is also a matched pair of brackets.
Given n strings of brackets, determine whether each sequence of 
brackets is balanced. If a string is balanced, return YES. Otherwise, return NO.

INPUT:

The first line contains a single integer n, the number of strings. 

Each of the next n lines contains a single string s, a sequence of brackets.

CONSTRAINTS:

1<=n<=10^3
1<=|s|<=10^3, where  is the length of the sequence.
All chracters in the sequences ? { {, }, (, ), [, ] }.

OUTPUT:

For each string, return YES or NO.
 

Sample Input
3
{[()]}
{[(])}
{{[[(())]]}}
Sample Output
YES
NO
YES
Time Limit: 5
Memory Limit: 256
Source Limit:
Explanation

1.The string {[()]} meets both criteria for being a balanced string, so we print YES on a new line.

2.The string {[(])} is not balanced because the brackets enclosed by the matched pair { and } are not balanced: [(]).

3.The string {{[[(())]]}} meets both criteria for being a balanced string, so we print YES on a new line.

Code:-

#include<stdio.h>
#include<stdlib.h>
struct stack
{
int top;
int capacity;
char *array;
};
struct stack* creatstack(int cap)
{
struct stack *stack;
stack=(struct stack*)malloc(sizeof(struct stack));
stack->top=-1;
stack->capacity=cap;
stack->array=(char *)malloc(sizeof(char)*stack->capacity);
return(stack);
}
int isfull(struct stack* stack)
{
if(stack->top==stack->capacity1)
return(0);
else
return(1);
}
int isempty(struct stack *stack)
{
if(stack->top==-1)
return(0);
else
return(1);
}
void push(struct stack *stack,char item)
{
if(isfull(stack))
{
stack->top++;
stack->array[stack->top]=item;
}
}
char pop(struct stack *stack)
{
char item;
if(isempty(stack))
{
item=stack->array[stack->top];
stack->top–;
return(item);
}
else
return(‘a’);
}
int main()
{
int i,flag=0,n,r,j;
char s[1000],ch;
struct stack* stack;
stack=creatstack(1000);
scanf(“%d”,&n);
for(j=1;j<=n;j++)
{
scanf(“%s”,s);
for(i=0;s[i];i++)
{
if(s[i]==‘(‘||s[i]==‘{‘||s[i]==‘[‘)
push(stack,s[i]);
if(s[i]==‘)’||s[i]==‘}’||s[i]==‘]’)
{
ch=pop(stack);
if(s[i]==‘)’)
{
if(ch!=‘(‘)
{
printf(“NOn”);
flag=1;
break;
}
}
if(s[i]==‘}’)
{
if(ch!=‘{‘)
{
printf(“NOn”);
flag=1;
break;
}
}
if(s[i]==‘]’)
{
if(ch!=‘[‘)
{
printf(“NOn”);
flag=1;
break;
}
}
}
}
r=isempty(stack);
if(r==0&&flag==0)
printf(“YESn”);
else
{
if(flag==0)
printf(“NOn”);
}
flag=0;
while(isempty(stack))
pop(stack);
}
return 0;
}



Leave a Reply

Your email address will not be published.