(HackerRank) Write a Modular C Programming code to solve INFIX TO POSTFIX CONVERSION, Help Swaraj to write an modular c program to based on given algorithm.
Swaraj wants to learn how to make use of stacks data structure to convert the infix expression to postfix expression. He was discussing with his class teacher. The class teacher provided him an agorithm on the conversion process.
Help Swaraj to write a modular c program to based on given algorithm.
The algorithm is as follows:
Input Format
Infix expression.
Constraints
Expression may or may not have parantheses.
Output Format
Converted postfix expression.
Sample Input 0
a+b
Sample Output 0
Given Infix Expression: a+b
Postfix Expression: ab+
Sample Input 1
(2+3*5)-1*3
Sample Output 1
Given Infix Expression: (2+3*5)-1*3
Postfix Expression: 235*+13*-
Sample Input 2
a*b+c/d
Sample Output 2
Given Infix Expression: a*b+c/d
Postfix Expression: ab*cd/+
CODE:
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define SIZE 100
struct stack
{
char data[SIZE];
int top;
};
void push(struct stack *sptr, char num);
char pop(struct stack *sptr);
void postfix_conv(char infix[SIZE]);
int pre(char ch);
int main()
{
char infix[50];
scanf("%s",infix);
postfix_conv(infix);
return 0;
}
void push(struct stack *sptr, char num)
{
if(sptr->top==SIZE-1)
{
printf("Stack Overflow\n");
}
else
{
sptr->top++;
sptr->data[sptr->top]=num;
}
}
char pop(struct stack *sptr)
{
char num;
num=sptr->data[sptr->top];
sptr->top--;
return num;
}
char peek(struct stack *sptr)
{
char c;
c=sptr->data[sptr->top];
return c;
}
void postfix_conv(char infix[SIZE])
{
struct stack s, * sptr;
sptr=&s;
sptr->top=-1;
char postfix[SIZE],ch,elem;
int i=0,k=0;
push(sptr,'#');
while(infix[i]!='\0')
{
ch=infix[i];
if(ch=='(')
{
push(sptr,ch);
}
else if(isalnum(ch))
{
postfix[k++]=ch;
}
else if(ch==')')
{
while(sptr->data[sptr->top]!='(')
{
postfix[k++]=pop(sptr);
}
elem=pop(sptr);
}
else
{
while(pre(peek(sptr))>=pre(ch))
{
postfix[k++]=pop(sptr);
}
push(sptr,ch);
}
i++;
}
while(sptr->data[sptr->top]!='#')
{
postfix[k++]=pop(sptr);
}
postfix[k]='\0';
printf("Given Infix Expression: %s\nPostfix Expression: %s",infix,postfix);
}
int pre(char ch)
{
switch(ch)
{
case '#':return 0;
break;
case '(':return 1;
break;
case '+':
case '-':return 2;
break;
case '/':
case '*':
case '%':return 3;
break;
case '^':return 4;
break;
}
return 0;
}
OUTPUT
Congratulations, you passed the sample test case. Click the Submit Code button to run your code against all the test cases. Input (stdin) a+b Your Output (stdout) Given Infix Expression: a+b Postfix Expression: ab+ Expected Output Given Infix Expression: a+b Postfix Expression: ab+
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