Given a positive integer
Examples:
Input : N = 2 Output : 1
Input : N = 20 Output : 3
Simple Approach: As per question a very simple and brute force approach is to iterate over N until it got reduced to 1, where reduction involve two cases:
- N is power of 2 : reduce n to n/2
- N is not power of 2: reduce n to n – (2^log2(n))
Efficient approach: Before proceeding to actual result lets have a look over bit representation of an integer n as per problem statement.
- When an integer is power of 2: In this case bit -representation includes only one set bit and that too is left most. Hence log2(n) i.e. bit-position minus One is the number of step required to reduce it to n. Which is also equal to number of set bit in n-1.
- When an integer is not power of 2:The remainder of n – 2^(log2(n)) is equal to integer which can be obtained by un-setting the left most set bit. Hence, one set bit removal count as one step in this case.
Hence the actual answer for steps required to reduce n is equal to number of set bits in n-1. Which can be easily calculated either by using the loop or any of method described in the post: Count Set bits in an Integer.
Below is the implementation of the above approach:
// Cpp to find the number of step to reduce n to 1 // C++ program to demonstrate __builtin_popcount() #include <iostream> using namespace std;
// Function to return number of steps for reduction int stepRequired( int n)
{ // builtin function to count set bits
return __builtin_popcount(n - 1);
} // Driver program int main()
{ int n = 94;
cout << stepRequired(n) << endl;
return 0;
} |
// Java program to find the number of step to reduce n to 1 import java.io.*;
class GFG
{ // Function to return number of steps for reduction
static int stepRequired( int n)
{
// builtin function to count set bits
return Integer.bitCount(n - 1 );
}
// Driver program
public static void main(String []args)
{
int n = 94 ;
System.out.println(stepRequired(n));
}
} // This code is contributed by // ihritik |
# Python3 to find the number of step # to reduce n to 1 # Python3 program to demonstrate # __builtin_popcount() # Function to return number of # steps for reduction def stepRequired(n) :
# step to count set bits
return bin ( 94 ).count( '1' )
# Driver Code if __name__ = = "__main__" :
n = 94
print (stepRequired(n))
# This code is contributed by Ryuga |
// C# program to find the number of step to reduce n to 1 using System;
class GFG
{ // function to count set bits
static int countSetBits( int n)
{
// base case
if (n == 0)
return 0;
else
// if last bit set
// add 1 else add 0
return (n & 1) + countSetBits(n >> 1);
}
// Function to return number of steps for reduction
static int stepRequired( int n)
{
return countSetBits(n - 1);
}
// Driver program
public static void Main()
{
int n = 94;
Console.WriteLine(stepRequired(n));
}
} // This code is contributed by // ihritik |
<?php // PHP program to find the number of step to reduce n to 1 // recursive function to // count set bits function countSetBits( $n )
{ // base case
if ( $n == 0)
return 0;
else
return 1 +
countSetBits( $n &
( $n - 1));
} // Function to return number of steps for reduction function stepRequired( $n )
{ return countSetBits( $n - 1);
} // Driver program $n = 94;
echo stepRequired( $n );
// This code is contributed by // ihritik ?> |
<script> // Javascript program to find the number of step to reduce n to 1
// function to count set bits
function countSetBits(n)
{
// base case
if (n == 0)
return 0;
else
// if last bit set
// add 1 else add 0
return (n & 1) + countSetBits(n >> 1);
}
// Function to return number of steps for reduction
function stepRequired(n)
{
return countSetBits(n - 1);
}
let n = 94;
document.write(stepRequired(n));
// This code is contributed by decode2207. </script> |
Output:
5
Time Complexity: O(1) as constant time is taken.
Auxiliary Space: O(1)