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Print Bracket Number

  • Difficulty Level : Easy
  • Last Updated : 21 May, 2021

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.

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Approach
 



  1. Define a variable left_bnum = 1.
  2. Create a stack right_bnum.
  3. 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
?>

Javascript




<script>
 
// Javascript 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
    var left_bnum = 1;
     
    // used to obtain the bracket number
    // for the right bracket
    var right_bnum = [];
     
    // traverse the given expression 'exp'
    for (var i = 0; i < n; i++) {
         
        // if current character is a left bracket
        if (exp[i] == '(') {
            // print 'left_bnum',
            document.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
            document.write( right_bnum[right_bnum.length-1] + " ");
             
            // pop the top element from the stack
            right_bnum.pop();
        }
    }
}
 
// Driver program to test above
var exp = "(a+(b*c))+(d/e)";
var n = exp.length;
 
printBracketNumber(exp, n);
 
</script>
Output: 
1 2 2 1 3 3

 

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

 




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