Minimize steps to reach K from 0 by adding 1 or doubling at each step

Given a positive integer K, the task is to find the minimum number of operations of the following two types, required to change 0 to K: 

• Add one to the operand
• Multiply the operand by 2.

Examples:

Input: K = 1 
Output: 1 
Explanation:
Step 1: 0 + 1 = 1 = K
Input: K = 4 
Output: 3 
Explanation:
Step 1: 0 + 1 = 1, 
Step 2: 1 * 2 = 2, 
Step 3: 2 * 2 = 4 = K 

Approach:

• If K is an odd number, the last step must be adding 1 to it.
• If K is an even number, the last step is to multiply by 2 to minimize the number of steps.
• Create a dp[] table that stores in every dp[i], the minimum steps required to reach i. 
  

Proof:

• Let us consider an even number X
• Now we have two options:
  i) X-1
  ii)X/2
• Now if we choose X-1:
  -> X-1 is an odd number , since X is even
  -> Hence we choose subtract 1  operation and we have X-2
  ->Now we have X-2, which is also an even number , Now again we have two options either to subtract 1 or to divide by 2
  ->Now let us choose divide by 2 operation , hence we have ( X – 2 )/2 => (X/2) -1
• Now if we choose X/2:
  -> We can do the subtract 1 operation to reach (X/2)-1
• Now let us consider sequence of both the cases:
  When we choose X-1 : X -> X-1 -> X-2 -> (X/2)-1 [ totally three operations ]
  When we choose X/2 : X -> X/2 -> (X/2)-1 [ totally two operations ]
  

This diagram shows how choosing divide operation for even numbers leads to optimal solution recursively

• Hence we can say that for a given even number , choosing the divide by 2 operation will always give us the minimum number of steps
• Hence proved

Below is the implementation of the above approach: 

C++

// C++ program to implement the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find minimum operations
int minOperation(int k)
{
    // vector dp is initialised
    // to store the steps
    vector<int> dp(k + 1, 0);

    for (int i = 1; i <= k; i++) {

        dp[i] = dp[i - 1] + 1;

        // For all even numbers
        if (i % 2 == 0) {
            dp[i]
                = min(dp[i],
                      dp[i / 2] + 1);
        }
    }
    return dp[k];
}

// Driver Code
int main()
{
    int K = 12;
    cout << minOperation(k);
}

Java

// Java program to implement the above approach
class GFG{

// Function to find minimum operations
static int minOperation(int k)
{

    // dp is initialised
    // to store the steps
    int dp[] = new int[k + 1];

    for(int i = 1; i <= k; i++)
    {
       dp[i] = dp[i - 1] + 1;

       // For all even numbers
       if (i % 2 == 0)
       {
           dp[i] = Math.min(dp[i], dp[i / 2] + 1);
       }
    }
    return dp[k];
}

// Driver Code
public static void main (String []args)
{
    int K = 12;
    System.out.print( minOperation(K));
}
}

// This code is contributed by chitranayal

Python3

# Python3 program to implement the above approach

# Function to find minimum operations
def minOperation(k):

    # dp is initialised
    # to store the steps
    dp = [0] * (k + 1)

    for i in range(1, k + 1):
        dp[i] = dp[i - 1] + 1

        # For all even numbers
        if (i % 2 == 0):
            dp[i]= min(dp[i], dp[i // 2] + 1)

    return dp[k]

# Driver Code
if __name__ == '__main__':

    k = 12

    print(minOperation(k))

# This code is contributed by mohit kumar 29

C#

// C# program to implement the above approach
using System;
class GFG{

// Function to find minimum operations
static int minOperation(int k)
{

    // dp is initialised
    // to store the steps
    int []dp = new int[k + 1];

    for(int i = 1; i <= k; i++)
    {
        dp[i] = dp[i - 1] + 1;

        // For all even numbers
        if (i % 2 == 0)
        {
            dp[i] = Math.Min(dp[i], dp[i / 2] + 1);
        }
    }
    return dp[k];
}

// Driver Code
public static void Main()
{
    int K = 12;
    Console.Write(minOperation(K));
}
}

// This code is contributed by Nidhi_Biet

Javascript

<script>
// Javascript implementation of the above approach

// Function to find minimum operations
function minOperation(k)
{

    // dp is initialised
    // to store the steps
    let dp = Array.from({length: k+1}, (_, i) => 0);

    for(let i = 1; i <= k; i++)
    {
       dp[i] = dp[i - 1] + 1;

       // For all even numbers
       if (i % 2 == 0)
       {
           dp[i] = Math.min(dp[i], dp[i / 2] + 1);
       }
    }
    return dp[k];
}

  // Driver Code
    let K = 12;
    document.write( minOperation(K));

 // This code is contributed by target_2.
</script>

Output:

5

Time Complexity: O(k)
Auxiliary Space: O(k)