Check if concatenation of two strings is balanced or not

Given two bracket sequences S1 and S2 consisting of ‘(‘ and ‘)’. The task is to check if the string obtained by concatenating both the sequences is balanced or not. Concatenation can be done by s1+s2 or s2+s1.

Examples:

Input: s1 = “)()(())))”, s2 = “(()(()(”
Output: Balanced
s2 + s1 = “(()(()()()(())))”, which
is a balanced paranthesis sequence.

Input: s1 = “(()))(“, s2 = “())())”
Output: Not balanced
s1 + s2 = “(()))(())())” –> Not balanced
s2 + s1 = “())())(()))(” –> Not balanced



A naive solution is to first concatenate both sequences and then check if the resultant sequence is balanced or not using a stack. First, check if s1 + s2 is balanced or not. If not, then check if s2 + s1 is balanced or not. To check if a given sequence of brackets is balanced or not using a stack, the following algorithm can be used.

  1. Declare a character stack S.
  2. Now traverse the expression string exp.
    • If the current character is a starting bracket (‘(‘ or ‘{‘ or ‘[‘) then push it to stack.
    • If the current character is a closing bracket (‘)’ or ‘}’ or ‘]’) then pop from the stack and if the popped character is the matching starting bracket then fine else parenthesis are not balanced.
  3. After complete traversal, if there is some starting bracket left in stack then “not balanced”.

Below is the implementation of above approach:

C++

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// CPP program to check if sequence obtained
// by concatenating two bracket sequences
// is balanced or not.
#include <bits/stdc++.h>
using namespace std;
  
// Check if given string is balanced bracket
// sequence or not.
bool isBalanced(string s)
{
  
    stack<char> st;
  
    int n = s.length();
  
    for (int i = 0; i < n; i++) {
  
        // If current bracket is an opening
        // bracket push it to stack.
        if (s[i] == '(')
            st.push(s[i]);
  
        // If current bracket is a closing
        // bracket then pop from stack if
        // it is not empty. If stack is empty
        // then sequence is not balanced.
        else {
            if (st.empty()) {
                return false;
            }
  
            else
                st.pop();
        }
    }
  
    // If stack is not empty, then sequence
    // is not balanced.
    if (!st.empty())
        return false;
  
    return true;
}
  
// Function to check if string obtained by
// concatenating two bracket sequences is
// balanced or not.
bool isBalancedSeq(string s1, string s2)
{
  
    // Check if s1 + s2 is balanced or not.
    if (isBalanced(s1 + s2))
        return true;
  
    // Check if s2 + s1 is balanced or not.
    return isBalanced(s2 + s1);
}
  
// Driver code.
int main()
{
    string s1 = ")()(())))";
    string s2 = "(()(()(";
  
    if (isBalancedSeq(s1, s2))
        cout << "Balanced";
    else
        cout << "Not Balanced";
  
    return 0;
}

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Output:

Balanced

Time complexity: O(n)
Auxiliary Space: O(n)

An efficient solution is to check if given sequences can result in balanced parenthesis sequence without using a stack, i.e., in constant extra space.
Let the concatenated sequence is s. There are two possibilities: either s = s1 + s2 is balanced or s = s2 + s1 is balanced. Check for both possibilities whether s is balanced or not.

  • If s is balanced, then the number of opening brackets in s should always be greater than or equal to the number of closing brackets in S at any instant of traversing it. This is because if at any instant number of closing brackets in s is greater than the number of opening brackets, then the last closing bracket will not have a matching opening bracket (that is why the count is more) in s.
  • If the sequence is balanced then at the end of traversal, the number of opening brackets in s is equal to the number of closing brackets in s.

Below is the implementation of above approach:

C++

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// C++ program to check if sequence obtained
// by concatenating two bracket sequences
// is balanced or not.
#include <bits/stdc++.h>
using namespace std;
  
// Check if given string is balanced bracket
// sequence or not.
bool isBalanced(string s)
{
  
    // To store result of comparison of
    // count of opening brackets and
    // closing brackets.
    int cnt = 0;
  
    int n = s.length();
  
    for (int i = 0; i < n; i++) {
  
        // If current bracket is an
        // opening bracket, then
        // increment count.
        if (s[i] == '(')
            cnt++;
  
        // If current bracket is a
        // closing bracket, then
        // decrement count and check
        // if count is negative.
        else {
            cnt--;
            if (cnt < 0)
                return false;
        }
    }
  
    // If count is positive then
    // some opening brackets are
    // not balanced.
    if (cnt > 0)
        return false;
  
    return true;
}
  
// Function to check if string obtained by
// concatenating two bracket sequences is
// balanced or not.
bool isBalancedSeq(string s1, string s2)
{
  
    // Check if s1 + s2 is balanced or not.
    if (isBalanced(s1 + s2))
        return true;
  
    // Check if s2 + s1 is balanced or not.
    return isBalanced(s2 + s1);
}
  
// Driver code.
int main()
{
    string s1 = ")()(())))";
    string s2 = "(()(()(";
  
    if (isBalancedSeq(s1, s2))
        cout << "Balanced";
    else
        cout << "Not Balanced";
  
    return 0;
}

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Java

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// Java program to check if 
// sequence obtained by
// concatenating two bracket 
// sequences is balanced or not.
import java.io.*;
  
class GFG
{
  
// Check if given string
// is balanced bracket 
// sequence or not.
static boolean isBalanced(String s)
{
      
// To store result of comparison 
// of count of opening brackets 
// and closing brackets.
int cnt = 0;
int n = s.length();
for (int i = 0; i < n; i++) 
{
      
    // If current bracket is 
    // an opening bracket, 
    // then increment count.
    if (s.charAt(i) =='(')
    {
        cnt = cnt + 1;
    }
      
    // If current bracket is a
    // closing bracket, then
    // decrement count and check
    // if count is negative.
    else
    {
        cnt = cnt - 1;
        if (cnt < 0)
            return false;
    }
}
  
// If count is positive then
// some opening brackets are
// not balanced.
if (cnt > 0)
    return false;
  
return true;
}
  
// Function to check if string 
// obtained by concatenating 
// two bracket sequences is
// balanced or not.
static boolean isBalancedSeq(String s1, 
                             String s2)
{
  
// Check if s1 + s2 is
// balanced or not.
if (isBalanced(s1 + s2))
    return true;
  
// Check if s2 + s1 is
// balanced or not.
return isBalanced(s2 + s1);
}
  
// Driver code
public static void main(String [] args)
{
    String s1 = ")()(())))";
    String s2 = "(()(()(";
      
    if (isBalancedSeq(s1, s2))
    {
        System.out.println("Balanced");
    }
    else
    
        System.out.println("Not Balanced");
    }
}
}
  
// This code is contributed 
// by Shivi_Aggarwal

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Python3

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# Python3 program to check 
# if sequence obtained by
# concatenating two bracket 
# sequences is balanced or not.
  
# Check if given string 
# is balanced bracket 
# sequence or not.
def isBalanced(s):
      
    # To store result of 
    # comparison of count 
    # of opening brackets 
    # and closing brackets.
    cnt = 0
    n = len(s)
  
    for i in range(0, n):
        if (s[i] == '('):
            cnt = cnt + 1
        else :
            cnt = cnt - 1
            if (cnt < 0):
                return False
    if (cnt > 0):
        return False
  
    return True
  
def isBalancedSeq(s1, s2):
  
    if (isBalanced(s1 + s2)):
        return True
  
    return isBalanced(s2 + s1)
  
# Driver code
a = ")()(())))";
b = "(()(()(";
  
if (isBalancedSeq(a, b)):
        print("Balanced")
else:
    print("Not Balanced")
  
# This code is contributed 
# by Shivi_Aggarwal

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Output:

Balanced

Time complexity: O(n)
Auxiliary Space: O(1)



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Improved By : Shivi_Aggarwal