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 minimise the number of steps.
  • Create a dp[] table that stores in every dp[i], the minimum steps required to reach i.

    dp[i] =\begin{cases} dp[i - 1] + 1 & \text{; if i is odd} \\ dp[\frac{i}{2}] + 1 & \text{; if i is even} \end{cases}

Below is the implementation of the above approach:

C++

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// 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);
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

5

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