Skip to content
Related Articles

Related Articles

Improve Article

Maximum steps to transform 0 to X with bitwise AND

  • Difficulty Level : Easy
  • Last Updated : 13 Jul, 2021

For an integer N, there are elements ranging from 0 to N-1. Certain elements can be transformed into other elements. Each transformation requires a certain effort which is equal to 1 unit, for each transformation. An element A can be transformed to element B, if and only if A != B and A & B = A (where & is the bitwise AND operator). We need to find the maximum effort possible to obtain the element with value X,  from the element with value 0, by a series of transformations.
Examples : 
 

Input : X = 2 
Output : 1 
The only way of obtaining 2 is to directly transform 0 to 2 (bitwise AND of 0 and 2 is 0) and hence requires one step. 
Input : X = 3 
Output : 2 
3 can be obtained in two steps. First, transform 0 to 1 (bitwise AND of 0 and 1 is 0). Then, transform 1 to 3 (bitwise AND of 1 and 3 is 1).

The simple solution is to count the number of set bits in X. 
Explanation: 
First, consider a single step transformation from A to B. All the set bits (bits which are equal to 1) of A should be set in B, otherwise bitwise AND of A and B will not be equal to A. If there is any bit that is set in A but not in B, then the transformation is not possible. The unset bits of A can either be set or unset in B, it does not matter. We can then change A to B by setting all the unset bits in A in one single step. Consequently, if we had to transform 0 to X in the least steps, the answer would have been one because bitwise AND of 0 with any number is 0. 
But we have to compute the maximum steps. So in each step, we set each bit starting from the right and a set bit can not be cleared once it set. 
Example: 
Suppose we want to obtain 13 (1101 in binary) from 0. We start by setting the 1st bit from the right by transforming 0 to 1 (0001 in binary). We next set the 3rd bit from the right to form 5 (0101 in binary). The last step would be to set the 4th bit and obtain 13 (1101).
 

C++




// CPP code to find the maximum possible
// effort
#include <bits/stdc++.h>
using namespace std;
 
// Function to get no of set bits in binary
// representation of positive integer n
unsigned int countSetBits(unsigned int n)
{
    unsigned int count = 0;
    while (n) {
        count += n & 1;
        n >>= 1;
    }
    return count;
}
 
// Driver code
int main()
{
    int i = 3;
    cout << countSetBits(i);
    return 0;
}

Java




// Java code to find the maximum
// possible effort
 
class GFG {
     
// Function to get no. of
// set bits in binary
// representation of
// positive integer n
static int countSetBits(int n)
{
    int count = 0;
    while (n != 0)
    {
        count += n & 1;
        n >>= 1;
    }
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    int i = 3;
    System.out.print(countSetBits(i));
}
}
 
// This code is contributed by Smitha.

Python3




# Python3 code to find the
# maximum possible effort
   
# Function to get no of
# set bits in binary
# representation of positive
# integer n
def countSetBits(n) :
    count = 0
    while (n) :
        count += n & 1
        n >>= 1
    return count
   
# Driver code
i = 3
print (countSetBits(i))
   
# This code is contributed by
# Manish Shaw(manishshaw1)

C#




// C# code to find the maximum
// possible effort
using System;
 
class GFG {
     
// Function to get no. of
// set bits in binary
// representation of
// positive integer n
static int countSetBits(int n)
{
    int count = 0;
    while (n != 0)
    {
        count += n & 1;
        n >>= 1;
    }
    return count;
}
 
// Driver code
public static void Main(String[] args)
{
    int i = 3;
    Console.Write(countSetBits(i));
}
}
 
// This code is contributed by Smitha.

PHP




<?php
// PHP code to find the
// maximum possible effort
 
// Function to get no of
// set bits in binary
// representation of positive
// integer n
function countSetBits($n)
{
    $count = 0;
    while ($n)
    {
        $count += $n & 1;
        $n >>= 1;
    }
    return $count;
}
 
// Driver code
$i = 3;
echo (countSetBits($i));
 
// This code is contributed by
// Manish Shaw(manishshaw1)
?>

Javascript




<script>
  
// JavaScript code to find the maximum possible
// effort
 
// Function to get no of set bits in binary
// representation of positive integer n
function countSetBits(n)
{
    var count = 0;
    while (n) {
        count += n & 1;
        n >>= 1;
    }
    return count;
}
 
// Driver code
var i = 3;
document.write( countSetBits(i));
 
</script>

Output:  

2

Time Complexity: O(log10N)

Auxiliary Space: O(1)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :