C Program to Check Expression is Correctly Parenthesized

This is a C Program to check if expression is correctly parenthesized.

Problem Description

This program takes a expression as input and checks if the expression is correctly parenthesized.

Problem Solution

1. Take a expression as input and store it in the array.
2. Check for the “(” and “)” in the expression.
3. If “(” encounters, then push it to the separate array. If “)” encounters, then pop the element of the array.
4. If the number of “(” and “)” are equal, then the expression is correctly parenthesized. Otherwise it is not.

Program/Source Code

Here is source code of the C Program to check if expression is correctly parenthesized. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

```
/*
```

 * C Program to Check if Expression is correctly Parenthesized

```
 */
```
```
#include <stdio.h>
```
```
#include <stdlib.h>
```
```
#include <string.h>
```
```
 
```
```
int top = -1;
```
```
char stack[100];
```
```
 
```
```
// function prototypes
```
```
void push(char);
```
```
void pop();
```
```
void find_top();
```
```
 
```
```
void main()
```
```
{
```
```
	int i;
```
```
	char a[100];
```
```
	printf("enter expression\n");
```
```
	scanf("%s", &a);
```
```
	for (i = 0; a[i] != '\0';i++)
```
```
	{
```
```
		if (a[i] == '(')
```
```
		{
```
```
			push(a[i]);
```
```
		}
```
```
		else if (a[i] == ')')
```
```
		{
```
```
			pop();
```
```
		}
```
```
	}
```
```
	find_top();
```
```
}
```
```
 
```
```
// to push elements in stack
```
```
void push(char a)
```
```
{
```
```
	stack[top] = a;
```
```
	top++;
```
```
}
```
```
 
```
```
// to pop elements from stack
```
```
void pop()
```
```
{
```
```
	if (top == -1)
```
```
	{
```
```
		printf("expression is invalid\n");
```
```
		exit(0);
```
```
	}	
```
```
	else
```
```
	{		
```
```
		top--;
```
```
	}
```
```
}
```
```
 
```
```
// to find top element of stack
```
```
void find_top()
```
```
{
```
```
	if (top == -1)
```
```
		printf("\nexpression is valid\n");
```
```
	else
```
```
		printf("\nexpression is invalid\n");
```
```
}
```

Program Explanation

1. Take a expression as input and store it in the array a[]. Initialize the variable top to -1.
2. Using for loop check for the “(” and “)” in the expression.
3. If “(” encounters, then push the character to the separate array stack[] and increment the variable top by 1. If “)” encounters, then pop the top most element of the array stack[] and decrement the variable top by 1. Use variable top to denote the top element of the array stack[].
4. If the variable top is equal to -1, then the expression is correctly parenthesized. Otherwise it is not.

Runtime Test Cases

enter expression
(a+b)
expression is valid
 
 
enter expression
(a+b))
expression is invalid
 
 
enter expression
((a+b)
expression is invalid
 
 
enter expression
((a+b)*(c+d))
expression is valid

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