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.

  • If N ≤ 0 or N ≥ MAX means it can not be converted to M so return MAX.
  • If N = M then return 0 as N got converted to M.
  • Else find the value at dp[N] if it is not -1, it means it has been calculated earlier so return dp[N].
  • If it is -1 then will call the recursive function as 2 * N and N – 1 and return the the minimum because if N is odd then it can be reached only by performing N – 1 operations and if N is even then 2 * N opearations have to be performed so check both the possibililties and return the minimum.

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

2

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