Convert N to M with given operations using dynamic programming

Given two integers N and M and the task is to convert N to M with the following operations:

  1. Multiply N by 2 i.e. N = N * 2.
  2. Subtract 1 from N i.e. N = N – 1.

Examples:

Input: N = 4, M = 6
Output: 2
Perform operation 2: N = N – 1 = 4 – 1 = 3
Perform operation 1: N = N * 2 = 3 * 2 = 6

Input: N = 10, M = 1
Output: 9

Approach: Create an array dp[] of size MAX = 105 + 5 to store the answer in order to prevent same computation again and again and initialize all the array elements with -1.



Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
const int N = 1e5 + 5;
  
int n, m;
int dp[N];
  
// Function to reutrn the minimum
// number of given operations
// required to convert n to m
int minOperations(int k)
{
    // If k is either 0 or out of range
    // then return max
    if (k <= 0 || k >= 2e4) {
        return 1e9;
    }
  
    // If k = m then conversion is
    // complete so return 0
    if (k == m) {
        return 0;
    }
  
    int& ans = dp[k];
  
    // If it has been calculated earlier
    if (ans != -1) {
        return ans;
    }
    ans = 1e9;
  
    // Call for 2*k and k-1 and return
    // the minimum of them. If k is even
    // then it can be reached by 2*k opertaions
    // and If k is odd then it can be reached
    // by k-1 opertaions so try both cases
    // and return the minimum of them
    ans = 1 + min(minOperations(2 * k),
                  minOperations(k - 1));
    return ans;
}
  
// Driver code
int main()
{
    n = 4, m = 6;
    memset(dp, -1, sizeof(dp));
  
    cout << minOperations(n);
  
    return 0;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach 
import java.util.*;
  
class GFG
{
    static final int N = 10000;
    static int n, m;
    static int[] dp = new int[N];
  
    // Function to reutrn the minimum
    // number of given operations
    // required to convert n to m
    static int minOperations(int k) 
    {
  
        // If k is either 0 or out of range
        // then return max
        if (k <= 0 || k >= 10000)
            return 1000000000;
  
        // If k = m then conversion is
        // complete so return 0
        if (k == m)
            return 0;
  
        dp[k] = dp[k];
  
        // If it has been calculated earlier
        if (dp[k] != -1)
            return dp[k];
        dp[k] = 1000000000;
  
        // Call for 2*k and k-1 and return
        // the minimum of them. If k is even
        // then it can be reached by 2*k opertaions
        // and If k is odd then it can be reached
        // by k-1 opertaions so try both cases
        // and return the minimum of them
        dp[k] = 1 + Math.min(minOperations(2 * k), 
                             minOperations(k - 1));
        return dp[k];
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        n = 4;
        m = 6;
        Arrays.fill(dp, -1);
        System.out.println(minOperations(n));
    }
}
  
// This code is contributed by
// sanjeev2552
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
N = 1000
dp = [-1] * N
  
# Function to reutrn the minimum
# number of given operations
# required to convert n to m
def minOperations(k):
  
    # If k is either 0 or out of range
    # then return max
    if (k <= 0 or k >= 1000): 
        return 1e9
      
    # If k = m then conversion is
    # complete so return 0
    if (k == m): 
        return 0
      
    dp[k] = dp[k]
      
    # If it has been calculated earlier
    if (dp[k] != -1): 
        return dp[k]
      
    dp[k] = 1e9
      
    # Call for 2*k and k-1 and return
    # the minimum of them. If k is even
    # then it can be reached by 2*k opertaions
    # and If k is odd then it can be reached
    # by k-1 opertaions so try both cases
    # and return the minimum of them
    dp[k] = 1 + min(minOperations(2 * k),
                    minOperations(k - 1))
    return dp[k]
  
# Driver code
if __name__ == '__main__':
    n = 4
    m = 6
    print(minOperations(n)) 
      
# This code is contributed by ashutosh450
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach 
using System;
using System.Linq;
  
class GFG
{
    static int N = 10000;
    static int n, m;
    static int[] dp = Enumerable.Repeat(-1, N).ToArray();
  
    // Function to reutrn the minimum
    // number of given operations
    // required to convert n to m
    static int minOperations(int k) 
    {
  
        // If k is either 0 or out of range
        // then return max
        if (k <= 0 || k >= 10000)
            return 1000000000;
  
        // If k = m then conversion is
        // complete so return 0
        if (k == m)
            return 0;
  
        dp[k] = dp[k];
  
        // If it has been calculated earlier
        if (dp[k] != -1)
            return dp[k];
        dp[k] = 1000000000;
  
        // Call for 2*k and k-1 and return
        // the minimum of them. If k is even
        // then it can be reached by 2*k opertaions
        // and If k is odd then it can be reached
        // by k-1 opertaions so try both cases
        // and return the minimum of them
        dp[k] = 1 + Math.Min(minOperations(2 * k), 
                             minOperations(k - 1));
        return dp[k];
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
        n = 4;
        m = 6;
          
        //Arrays.fill(dp, -1);
        Console.Write(minOperations(n));
    }
}
  
// This code is contributed by
// Mohit kumar 29
chevron_right

Output:
2

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.





Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :