Convert 0 to N by adding 1 or multiplying by 2 in minimum steps

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Given a positive integer N, the task is to find the minimum number of addition operations required to convert the number 0 to N such that in each operation any number can be multiplied by 2 or add the value 1 to it.

Examples:

Input: N = 6
Output: 1
Explanation:
Following are the operations performed to convert 0 to 6:
Add 1 –> 0 + 1 = 1.
Multiply 2 –> 1 * 2 = 2.
Add 1 –> 2 + 1 = 3.
Multiply 2 –> 3 * 2 = 6.
Therefore number of addition operations = 2.

Input: N = 3
Output: 2

Approach: This problem can be solved by using the Bit Manipulation technique. In binary number representation of N, while operating each bit whenever N becomes odd (that means the least significant bit of N is set) then perform the addition operation. Otherwise, multiply by 2. The final logic to the given problem is to find the number of set bits in N.

Below is the implementation of the above approach:

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count number of
// set bits in N
int minimumAdditionOperation(
    unsigned long long int N)
{
 
    // Stores the count of set bits
    int count = 0;
 
    while (N) {
 
        // If N is odd, then it
        // a set bit
        if (N & 1 == 1) {
            count++;
        }
        N = N >> 1;
    }
 
    // Return the result
    return count;
}
 
// Driver Code
int main()
{
    int N = 6;
    cout << minimumAdditionOperation(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG {
 
    // Function to count number of
    // set bits in N
    static int minimumAdditionOperation(int N)
    {
 
        // Stores the count of set bits
        int count = 0;
 
        while (N > 0) {
 
            // If N is odd, then it
            // a set bit
            if (N % 2 == 1) {
                count++;
            }
            N = N >> 1;
        }
 
        // Return the result
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 6;
        System.out.println(minimumAdditionOperation(N));
    }
}
 
// This code is contributed by dwivediyash

Python3

# python program for above approach
 
# Function to count number of
# set bits in N
def minimumAdditionOperation(N):
 
    # Stores the count of set bits
    count = 0
 
    while (N):
 
        # If N is odd, then it
        # a set bit
        if (N & 1 == 1):
            count += 1
 
        N = N >> 1
 
    # Return the result
    return count
 
# Driver Code
if __name__ == "__main__":
 
    N = 6
    print(minimumAdditionOperation(N))
 
    # This code is contributed by rakeshsahni.

C#

// C# program for above approach
using System;
public class GFG{
     
    // Function to count number of
    // set bits in N
    static int minimumAdditionOperation(int N)
    {
     
        // Stores the count of set bits
        int count = 0;
     
        while (N != 0) {
     
            // If N is odd, then it
            // a set bit
            if ((N & 1) == 1) {
                count++;
            }
            N = N >> 1;
        }
     
        // Return the result
        return count;
    }
     
    // Driver Code
    static public void Main (){
        int N = 6;
        Console.Write(minimumAdditionOperation(N));
     
    }
}
 
// This code is contributed by AnkThon

Javascript

<script>
// Javascript program for above approach
 
// Function to count number of
// set bits in N
function minimumAdditionOperation(N)
{
 
  // Stores the count of set bits
  let count = 0;
 
  while (N) 
  {
   
    // If N is odd, then it
    // a set bit
    if (N & (1 == 1)) {
      count++;
    }
    N = N >> 1;
  }
 
  // Return the result
  return count;
}
 
// Driver Code
let N = 6;
document.write(minimumAdditionOperation(N));
 
// This code is contributed by saurabh_jaiswal.
</script>

Output:

Time Complexity: O(log N)
Auxiliary Space: O(1)

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