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Check for balanced parenthesis without using stack
  • Difficulty Level : Medium
  • Last Updated : 23 Apr, 2021

Given an expression string exp, write a program to examine whether the pairs and the orders of “{“, ”}”, ”(“, ”)”, ”[“, ”]” are correct in exp. 
Examples: 

Input : exp = “[()]{}{[()()]()}”
Output : true

Input : exp = “[(])”
Output : false

We have discussed a stack based solution. Here we are not allowed to use the stack. Looks like this problem cannot be solved without extra space (please see comments at the end). We use recursion to solve the problem. 
Below is the implementation of the above algorithm:  

C++




// CPP program to check if parenthesis are
// balanced or not in an expression.
#include <bits/stdc++.h>
using namespace std;
 
char findClosing(char c)
{
    if (c == '(')
        return ')';
    if (c == '{')
        return '}';
    if (c == '[')
        return ']';
    return -1;
}
 
// function to check if parenthesis are
// balanced.
bool check(char expr[], int n)
{
    // Base cases
    if (n == 0)
        return true;
    if (n == 1)
        return false;
    if (expr[0] == ')' || expr[0] == '}' || expr[0] == ']')
        return false;
 
    // Search for closing bracket for first
    // opening bracket.
    char closing = findClosing(expr[0]);
 
    // count is used to handle cases like
    // "((()))".  We basically need to
    // consider matching closing bracket.
    int i, count = 0;
    for (i = 1; i < n; i++) {
        if (expr[i] == expr[0])
            count++;
        if (expr[i] == closing) {
            if (count == 0)
                break;
            count--;
        }
    }
 
    // If we did not find a closing
    // bracket
    if (i == n)
        return false;
 
    // If closing bracket was next
    // to open
    if (i == 1)
        return check(expr + 2, n - 2);
 
    // If closing bracket was somewhere
    // in middle.
    return check(expr + 1, i - 1) && check(expr + i + 1, n - i - 1);
}
 
// Driver program to test above function
int main()
{
    char expr[] = "[(])";
    int n = strlen(expr);
    if (check(expr, n))
        cout << "Balanced";
    else
        cout << "Not Balanced";
    return 0;
}

Java




// Java program to check if parenthesis are
// balanced or not in an expression.
import java.util.Arrays;
 
class GFG {
 
    static char findClosing(char c)
    {
        if (c == '(')
            return ')';
        if (c == '{')
            return '}';
        if (c == '[')
            return ']';
        return Character.MIN_VALUE;
    }
 
    // function to check if parenthesis are
    // balanced.
    static boolean check(char expr[], int n)
    {
        // Base cases
        if (n == 0)
            return true;
        if (n == 1)
            return false;
        if (expr[0] == ')' || expr[0] == '}' || expr[0] == ']')
            return false;
 
        // Search for closing bracket for first
        // opening bracket.
        char closing = findClosing(expr[0]);
 
        // count is used to handle cases like
        // "((()))". We basically need to
        // consider matching closing bracket.
        int i, count = 0;
        for (i = 1; i < n; i++) {
            if (expr[i] == expr[0])
                count++;
            if (expr[i] == closing) {
                if (count == 0)
                    break;
                count--;
            }
        }
 
        // If we did not find a closing
        // bracket
        if (i == n)
            return false;
 
        // If closing bracket was next
        // to open
        if (i == 1)
            return check(Arrays.copyOfRange(expr, i + 1, n), n - 2);
        // If closing bracket was somewhere
        // in middle.
        return check(Arrays.copyOfRange(expr, 1, n), i - 1) && check(Arrays.copyOfRange(expr, (i + 1), n), n - i - 1);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        char expr[] = "[(])".toCharArray();
        int n = expr.length;
        if (check(expr, n))
            System.out.println("Balanced");
        else
            System.out.println("Not Balanced");
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3




# Python3 program to check if parenthesis are
# balanced or not in an expression.
def findClosing(c):
    if c == '(':
        return ')'
    elif c == '{':
        return '}'
    elif c == '[':
        return ']'
    return -1
 
# function to check if parenthesis
# are balanced.
def check(expr, n):
 
    # Base cases
    if n == 0:
        return True
    if n == 1:
        return False
    if expr[0] == ')' or \
       expr[0] == '}' or expr[0] == ']':
        return False
 
    # Search for closing bracket for first
    # opening bracket.
    closing = findClosing(expr[0])
 
    # count is used to handle cases like
    # "((()))". We basically need to
    # consider matching closing bracket.
    i = -1
    count = 0
    for i in range(1, n):
        if expr[i] == expr[0]:
            count += 1
        if expr[i] == closing:
            if count == 0:
                break
            count -= 1
 
    # If we did not find a closing
    # bracket
    if i == n:
        return False
 
    # If closing bracket was next
    # to open
    if i == 1:
        return check(expr[2:], n - 2)
 
    # If closing bracket was somewhere
    # in middle.
    return check(expr[1:], i - 1) and \
           check(expr[i + 1:], n - i - 1)
 
# Driver Code
if __name__ == "__main__":
    expr = "[(])"
    n = len(expr)
 
    if check(expr, n):
        print("Balanced")
    else:
        print("Not Balanced")
 
# This code is conributed by
# sanjeev2552

C#




// C# program to check
// if parenthesis are
// balanced or not in
// an expression.
using System;
class GFG{
     
static char[] copyOfRange (char[] src,
                           int start,
                           int end)
{
  int len = end - start;
  char[] dest = new char[len];
  Array.Copy(src, start,
             dest, 0, len);
  return dest;
}
 
static char findClosing(char c)
{
  if (c == '(')
    return ')';
  if (c == '{')
    return '}';
  if (c == '[')
    return ']';
  return char.MinValue;
}
 
// Function to check if
// parenthesis are balanced.
static bool check(char []expr,
                  int n)
{
  // Base cases
  if (n == 0)
    return true;
  if (n == 1)
    return false;
  if (expr[0] == ')' ||
      expr[0] == '}' ||
      expr[0] == ']')
    return false;
 
  // Search for closing bracket for first
  // opening bracket.
  char closing = findClosing(expr[0]);
 
  // count is used to handle cases like
  // "((()))". We basically need to
  // consider matching closing bracket.
  int i, count = 0;
  for (i = 1; i < n; i++)
  {
    if (expr[i] == expr[0])
      count++;
    if (expr[i] == closing)
    {
      if (count == 0)
        break;
      count--;
    }
  }
 
  // If we did not find
  // a closing bracket
  if (i == n)
    return false;
 
  // If closing bracket
  // was next to open
  if (i == 1)
    return check(copyOfRange(expr,
                             i + 1, n),
                              n - 2);
  // If closing bracket
  // was somewhere in middle.
  return check(copyOfRange(expr, 1, n),
                           i - 1) &&
         check(copyOfRange(expr, (i + 1),
                           n), n - i - 1);
}
 
// Driver code
public static void Main(String[] args)
{
  char []expr = "[(])".ToCharArray();
  int n = expr.Length;
  if (check(expr, n))
    Console.WriteLine("Balanced");
  else
    Console.WriteLine("Not Balanced");
}
}
 
// This code is contributed by gauravrajput1

Javascript




<script>
 
// Javascript program to check if parenthesis are
// balanced or not in an expression.
 
function findClosing(c)
{
    if (c == '(')
        return ')';
    if (c == '{')
        return '}';
    if (c == '[')
        return ']';
    return -1;
}
 
// function to check if parenthesis are
// balanced.
function check(expr, n)
{
    // Base cases
    if (n == 0)
        return true;
    if (n == 1)
        return false;
    if (expr[0] == ')' || expr[0] == '}' || expr[0] == ']')
        return false;
 
    // Search for closing bracket for first
    // opening bracket.
    var closing = findClosing(expr[0]);
 
    // count is used to handle cases like
    // "((()))".  We basically need to
    // consider matching closing bracket.
    var i, count = 0;
    for (i = 1; i < n; i++) {
        if (expr[i] == expr[0])
            count++;
        if (expr[i] == closing) {
            if (count == 0)
                break;
            count--;
        }
    }
 
    // If we did not find a closing
    // bracket
    if (i == n)
        return false;
 
    // If closing bracket was next
    // to open
    if (i == 1)
        return check(expr + 2, n - 2);
 
    // If closing bracket was somewhere
    // in middle.
    return check(expr + 1, i - 1) && check(expr + i + 1, n - i - 1);
}
 
// Driver program to test above function
var expr = "[(])";
var n = expr.length;
if (check(expr, n))
    document.write( "Balanced");
else
    document.write( "Not Balanced");
 
// This code is contributed by itsok.
</script>
Output: 
Not Balanced

 

The above solution is very inefficient compared to the stack-based solution. This seems to only useful for recursion practice problems. 

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