Print steps to make a number in form of 2^X – 1
Given a number N, there are two steps to be performed.
- At odd step, XOR the number with any 2^M-1, where M is chosen by you.
- At even step, increase the number by 1.
Keep performing the steps until N becomes 2^X-1 (where x can be any integer). The task is to print all the steps.
Examples:
Input: 39
Output:
Step1: Xor with 31
Step2: Increase by 1
Step3: Xor with 7
Step4: Increase by 1
Pick M = 5, N is transformed into 39 ^ 31 = 56.
Increase N by 1 changing its value to N = 57.
Pick M = 3, x is transformed into 57 ^ 7 = 62.
Increase X by 1, changing its value to 63 = 2^6 – 1.
Input: 7
Output: No steps required.
Approach: The following steps can be followed to solve the above problem:
- At every odd step, find the leftmost unset bit (say position x in terms of 1 indexing) in the number and do a xor with 2^x-1.
- At every even step, increase the number by 1.
- If at any step, the number has no more unset bits, then return.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the leftmost// unset bit in a number.int find_leftmost_unsetbit(int n){ int ind = -1; int i = 1; while (n) { if (!(n & 1)) ind = i; i++; n >>= 1; } return ind;}// Function that perform// the stepvoid perform_steps(int n){ // Find the leftmost unset bit int left = find_leftmost_unsetbit(n); // If the number has no bit // unset, it means it is in form 2^x -1 if (left == -1) { cout << "No steps required"; return; } // Count the steps int step = 1; // Iterate till number is of form 2^x - 1 while (find_leftmost_unsetbit(n) != -1) { // At even step increase by 1 if (step % 2 == 0) { n += 1; cout << "Step" << step << ": Increase by 1\n"; } // Odd step xor with any 2^m-1 else { // Find the leftmost unset bit int m = find_leftmost_unsetbit(n); // 2^m-1 int num = pow(2, m) - 1; // Perform the step n = n ^ num; cout << "Step" << step << ": Xor with " << num << endl; } // Increase the steps step += 1; }}// Driver codeint main(){ int n = 39; perform_steps(n); return 0;} |
Java
// Java program to implement// the above approachclass GFG{ // Function to find the leftmost // unset bit in a number. static int find_leftmost_unsetbit(int n) { int ind = -1; int i = 1; while (n > 0) { if ((n % 2) != 1) { ind = i; } i++; n >>= 1; } return ind; } // Function that perform // the step static void perform_steps(int n) { // Find the leftmost unset bit int left = find_leftmost_unsetbit(n); // If the number has no bit // unset, it means it is in form 2^x -1 if (left == -1) { System.out.print("No steps required"); return; } // Count the steps int step = 1; // Iterate till number is of form 2^x - 1 while (find_leftmost_unsetbit(n) != -1) { // At even step increase by 1 if (step % 2 == 0) { n += 1; System.out.println("Step" + step + ": Increase by 1"); } // Odd step xor with any 2^m-1 else { // Find the leftmost unset bit int m = find_leftmost_unsetbit(n); // 2^m-1 int num = (int) (Math.pow(2, m) - 1); // Perform the step n = n ^ num; System.out.println("Step" + step + ": Xor with " + num); } // Increase the steps step += 1; } } // Driver code public static void main(String[] args) { int n = 39; perform_steps(n); }}// This code contributed by Rajput-Ji |
Python3
# Python3 program to implement# the above approach# Function to find the leftmost# unset bit in a number.def find_leftmost_unsetbit(n): ind = -1; i = 1; while (n): if ((n % 2) != 1): ind = i; i += 1; n >>= 1; return ind;# Function that perform# the stepdef perform_steps(n): # Find the leftmost unset bit left = find_leftmost_unsetbit(n); # If the number has no bit # unset, it means it is in form 2^x -1 if (left == -1): print("No steps required"); return; # Count the steps step = 1; # Iterate till number is of form 2^x - 1 while (find_leftmost_unsetbit(n) != -1): # At even step increase by 1 if (step % 2 == 0): n += 1; print("Step" , step , ": Increase by 1\n"); # Odd step xor with any 2^m-1 else: # Find the leftmost unset bit m = find_leftmost_unsetbit(n); # 2^m-1 num = (2**m) - 1; # Perform the step n = n ^ num; print("Step" , step, ": Xor with" , num ); # Increase the steps step += 1;# Driver coden = 39;perform_steps(n);# This code contributed by PrinciRaj1992 |
C#
// C# program to implement// the above approachusing System;class GFG{ // Function to find the leftmost // unset bit in a number. static int find_leftmost_unsetbit(int n) { int ind = -1; int i = 1; while (n > 0) { if ((n % 2) != 1) { ind = i; } i++; n >>= 1; } return ind; } // Function that perform // the step static void perform_steps(int n) { // Find the leftmost unset bit int left = find_leftmost_unsetbit(n); // If the number has no bit // unset, it means it is in form 2^x -1 if (left == -1) { Console.Write("No steps required"); return; } // Count the steps int step = 1; // Iterate till number is of form 2^x - 1 while (find_leftmost_unsetbit(n) != -1) { // At even step increase by 1 if (step % 2 == 0) { n += 1; Console.WriteLine("Step" + step + ": Increase by 1"); } // Odd step xor with any 2^m-1 else { // Find the leftmost unset bit int m = find_leftmost_unsetbit(n); // 2^m-1 int num = (int) (Math.Pow(2, m) - 1); // Perform the step n = n ^ num; Console.WriteLine("Step" + step + ": Xor with " + num); } // Increase the steps step += 1; } } // Driver code static public void Main () { int n = 39; perform_steps(n); }}// This code contributed by ajit. |
PHP
<?php// Php program to implement// the above approach// Function to find the leftmost// unset bit in a number.function find_leftmost_unsetbit($n){ $ind = -1; $i = 1; while ($n) { if (!($n & 1)) $ind = $i; $i++; $n >>= 1; } return $ind;}// Function that perform// the stepfunction perform_steps($n){ // Find the leftmost unset bit $left = find_leftmost_unsetbit($n); // If the number has no bit // unset, it means it is in form 2^x -1 if ($left == -1) { echo "No steps required"; return; } // Count the steps $step = 1; // Iterate till number is of form 2^x - 1 while (find_leftmost_unsetbit($n) != -1) { // At even step increase by 1 if ($step % 2 == 0) { $n += 1; echo "Step",$step, ": Increase by 1\n"; } // Odd step xor with any 2^m-1 else { // Find the leftmost unset bit $m = find_leftmost_unsetbit($n); // 2^m-1 $num = pow(2, $m) - 1; // Perform the step $n = $n ^ $num; echo "Step",$step ,": Xor with ", $num ,"\n"; } // Increase the steps $step += 1; }} // Driver code $n = 39; perform_steps($n); // This code is contributed by Ryuga?> |
Javascript
<script>// Javascript program to implement// the above approach// Function to find the leftmost// unset bit in a number.function find_leftmost_unsetbit(n){ let ind = -1; let i = 1; while (n) { if (!(n & 1)) ind = i; i++; n >>= 1; } return ind;}// Function that perform// the stepfunction perform_steps(n){ // Find the leftmost unset bit let left = find_leftmost_unsetbit(n); // If the number has no bit // unset, it means it is in form 2^x -1 if (left == -1) { document.write("No steps required"); return; } // Count the steps let step = 1; // Iterate till number is of form 2^x - 1 while (find_leftmost_unsetbit(n) != -1) { // At even step increase by 1 if (step % 2 == 0) { n += 1; document.write( "Step" + step + ": Increase by 1<br>" ); } // Odd step xor with any 2^m-1 else { // Find the leftmost unset bit let m = find_leftmost_unsetbit(n); // 2^m-1 let num = Math.pow(2, m) - 1; // Perform the step n = n ^ num; document.write("Step" + step + ": Xor with " + num + "<br>"); } // Increase the steps step += 1; }}// Driver code let n = 39; perform_steps(n);</script> |
Output:
Step1: Xor with 31 Step2: Increase by 1 Step3: Xor with 7 Step4: Increase by 1
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