Maximum steps to transform 0 to X with bitwise AND

For an integer N, there are elements ranging from 0 to N-1. Certain elements can be transformed to 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 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 which 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 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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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)

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C#

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// 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.

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PHP

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<?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)
?>

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Output :

2


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