Print Bracket Number
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.
Approach :
- Define a variable left_bnum = 1.
- Create a stack right_bnum.
- Now, for i = 0 to n-1.
- If exp[i] == ‘(‘, then print left_bnum, push left_bnum on to the stack right_bnum and finally increment left_bnum by 1.
- 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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