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Maximum steps to transform 0 to X with bitwise AND
• Difficulty Level : Easy
• Last Updated : 24 May, 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 ``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

 `>= 1;``    ``}``    ``return` `\$count``;``}` `// Driver code``\$i` `= 3;``echo` `(countSetBits(``\$i``));` `// This code is contributed by``// Manish Shaw(manishshaw1)``?>`

## Javascript

 ``

Output:

`2`

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