Print Bracket Number - GeeksforGeeks

Given an expression exp of length n consisting of some brackets. The task is to print the bracket numbers when the expression is being parsed.

Examples :

Input : (a+(b*c))+(d/e)
Output : 1 2 2 1 3 3
The highlighted brackets in the given expression
(a+(b*c))+(d/e) has been assigned the numbers as:
1 2 2 1 3 3.

Input : ((())(()))
Output : 1 2 3 3 2 4 5 5 4 1

Source: Flipkart Interview Experience | Set 49.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach :

Define a variable left_bnum = 1.
Create a stack right_bnum.
Now, for i = 0 to n-1.
1. If exp[i] == ‘(‘, then print left_bnum, push left_bnum on to the stack right_bnum and finally increment left_bnum by 1.
2. Else if exp[i] == ‘)’, then print the top element of the stack right_bnum and then pop the top element from the stack.

C++

// C++ implementation to print the bracket number 
#include <bits/stdc++.h> 
  
using namespace std; 
  
// function to print the bracket number 
void printBracketNumber(string exp, int n) 
{ 
    // used to print the bracket number 
    // for the left bracket 
    int left_bnum = 1; 
      
    // used to obtain the bracket number  
    // for the right bracket 
    stack<int> right_bnum; 
      
    // traverse the given expression 'exp' 
    for (int i = 0; i < n; i++) { 
          
        // if current character is a left bracket 
        if (exp[i] == '(') { 
            // print 'left_bnum', 
            cout << left_bnum << " "; 
              
            // push 'left_bum' on to the stack 'right_bnum' 
            right_bnum.push(left_bnum); 
              
            // increment 'left_bnum' by 1 
            left_bnum++; 
        } 
          
        // else if current character is a right bracket 
        else if(exp[i] == ')') { 
  
            // print the top element of stack 'right_bnum' 
            // it will be the right bracket number 
            cout << right_bnum.top() << " "; 
              
            // pop the top element from the stack 
            right_bnum.pop(); 
        } 
    } 
} 
  
// Driver program to test above 
int main() 
{ 
    string exp = "(a+(b*c))+(d/e)"; 
    int n = exp.size(); 
      
    printBracketNumber(exp, n); 
      
    return 0; 
} 

Java

// Java implementation to  
// print the bracket number 
import java.io.*; 
import java.util.*; 
  
class GFG 
{ 
    // function to print  
    // the bracket number 
    static void printBracketNumber(String exp,  
                                   int n) 
    { 
        // used to print the  
        // bracket number for  
        // the left bracket 
        int left_bnum = 1; 
          
        // used to obtain the  
        // bracket number for 
        // the right bracket 
        Stack<Integer> right_bnum = 
                   new Stack<Integer>(); 
          
        // traverse the given 
        // expression 'exp' 
        for (int i = 0; i < n; i++)  
        { 
              
            // if current character  
            // is a left bracket 
            if (exp.charAt(i) == '(')  
            { 
                  
                // print 'left_bnum', 
                System.out.print( 
                       left_bnum + " "); 
                  
                // push 'left_bum' on to 
                // the stack 'right_bnum' 
                right_bnum.push(left_bnum); 
                  
                // increment 'left_bnum' by 1 
                left_bnum++; 
            } 
              
            // else if current character  
            // is a right bracket 
            else if(exp.charAt(i) == ')') 
            { 
      
                // print the top element  
                // of stack 'right_bnum' 
                // it will be the right  
                // bracket number 
                System.out.print( 
                       right_bnum.peek() + " "); 
                  
                // pop the top element  
                // from the stack 
                right_bnum.pop(); 
            } 
        } 
    } 
      
    // Driver Code 
    public static void main(String args[]) 
    { 
        String exp = "(a+(b*c))+(d/e)"; 
        int n = exp.length(); 
          
        printBracketNumber(exp, n); 
    } 
} 
  
// This code is contributed  
// by Manish Shaw(manishshaw1) 

Python3

# Python3 implementation to print the bracket number 
  
# function to print the bracket number 
def printBracketNumber(exp, n): 
  
    # used to print the bracket number 
    # for the left bracket 
    left_bnum = 1
  
    # used to obtain the bracket number 
    # for the right bracket 
    right_bnum = list() 
  
    # traverse the given expression 'exp' 
    for i in range(n): 
  
        # if current character is a left bracket 
        if exp[i] == '(': 
  
            # print 'left_bnum', 
            print(left_bnum, end = " ") 
  
            # push 'left_bum' on to the stack 'right_bnum' 
            right_bnum.append(left_bnum) 
  
            # increment 'left_bnum' by 1 
            left_bnum += 1
  
        # else if current character is a right bracket 
        elif exp[i] == ')': 
  
            # print the top element of stack 'right_bnum' 
            # it will be the right bracket number 
            print(right_bnum[-1], end = " ") 
  
            # pop the top element from the stack 
            right_bnum.pop() 
  
# Driver Code 
if __name__ == "__main__": 
    exp = "(a+(b*c))+(d/e)"
    n = len(exp) 
  
    printBracketNumber(exp, n) 
  
# This code is contributed by 
# sanjeev2552 

C#

// C# implementation to  
// print the bracket number 
using System; 
using System.Collections.Generic; 
  
class GFG 
{ 
    // function to print  
    // the bracket number 
    static void printBracketNumber(string exp,  
                                   int n) 
    { 
        // used to print the bracket  
        // number for the left bracket 
        int left_bnum = 1; 
          
        // used to obtain the bracket   
        // number for the right bracket 
        Stack<int> right_bnum = new Stack<int>(); 
          
        // traverse the given 
        // expression 'exp' 
        for (int i = 0; i < n; i++)  
        { 
              
            // if current character  
            // is a left bracket 
            if (exp[i] == '(')  
            { 
                  
                // print 'left_bnum', 
                Console.Write(left_bnum + " "); 
                  
                // Push 'left_bum' on to 
                // the stack 'right_bnum' 
                right_bnum.Push(left_bnum); 
                  
                // increment 'left_bnum' by 1 
                left_bnum++; 
            } 
              
            // else if current character  
            // is a right bracket 
            else if(exp[i] == ')') 
            { 
      
                // print the top element  
                // of stack 'right_bnum' 
                // it will be the right  
                // bracket number 
                Console.Write(right_bnum.Peek() + " "); 
                  
                // Pop the top element  
                // from the stack 
                right_bnum.Pop(); 
            } 
        } 
    } 
      
    // Driver Code 
    static void Main() 
    { 
        string exp = "(a+(b*c))+(d/e)"; 
        int n = exp.Length; 
          
        printBracketNumber(exp, n); 
    } 
} 
  
// This code is contributed  
// by Manish Shaw(manishshaw1) 

PHP

<?php 
// PHP implementation to  
// print the bracket number 
  
// function to print 
// the bracket number 
function printBracketNumber($exp, $n) 
{ 
    // used to print the  
    // bracket number for  
    // the left bracket 
    $left_bnum = 1; 
      
    // used to obtain the  
    // bracket number for  
    // the right bracket 
    $right_bnum = array(); 
    $t = 0; 
      
    // traverse the given 
    // expression 'exp' 
    for ($i = 0; $i < $n; $i++)  
    { 
          
        // if current character  
        // is a left bracket 
        if ($exp[$i] == '(') 
        { 
              
            // print 'left_bnum', 
            echo $left_bnum . " "; 
              
            // push 'left_bum' on to 
            // the stack 'right_bnum' 
            $right_bnum[$t++] = $left_bnum; 
              
            // increment 'left_bnum' by 1 
            $left_bnum++; 
        } 
          
        // else if current character 
        // is a right bracket 
        else if($exp[$i] == ')')  
        { 
  
            // print the top element  
            // of stack 'right_bnum' 
            // it will be the right  
            // bracket number 
            echo $right_bnum[$t - 1] . " "; 
              
            // pop the top element 
            // from the stack 
            $right_bnum[$t - 1] = 1; 
            $t--; 
        } 
    } 
} 
  
// Driver Code 
$exp = "(a+(b*c))+(d/e)"; 
$n = strlen($exp); 
  
printBracketNumber($exp, $n); 
      
// This code is contributed  
// by mits 
?> 

Output:

1 2 2 1 3 3

Time Complexity : O(n).
Auxiliary Space : O(n).

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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