Check if N can be obtained from 1 by repetitively multiplying by 10 or 20

Given an integer N, the task is to determine whether it is possible to obtain the value N from 1 by repetitively multiplying by 10 or 20. Print Yes if possible or No otherwise.
Examples:

Input: N = 200
Output: YES
Explanation:
1 * 10 -> 10 * 20 -> 200
Input: N = 90
Output: NO

Approach:
Follow the steps below to solve the problem:

1. Count the number of trailing zeroes.
2. After removing the trailing zeroes, if the remaining N cannot be expressed as a power of 2, print NO.
3. Otherwise, if log2N <= Count of trailing zeroes, print Yes.

Below is the implementation of the above approach:

C++

 // C++ program to check if N  // can be obtained from 1 by  // repetitive multiplication  // by 10 or 20  #include using namespace std;    // Function to check if N can  // be obtained or not  void Is_possible(long long int N) {     int C = 0;     int D = 0;            // Count and remove trailing      // zeroes      while (N % 10 == 0)     {         N = N / 10;         C += 1;     }            // Check if remaining      // N is a power of 2      if(pow(2, (int)log2(N)) == N)      {         D = (int)log2(N);                    // To check the condition          // to print YES or NO          if (C >= D)             cout << "YES";         else             cout << "NO";      }      else         cout << "NO";  }    // Driver code int main() {     long long int N = 2000000000000;            Is_possible(N);  }    // This code is contributed by Stream_Cipher

Java

 // Java program to check if N  // can be obtained from 1 by  // repetitive multiplication  // by 10 or 20  import java.util.*;     class GFG{        static void Is_possible(long N) {     long C = 0;     long D = 0;            // Count and remove trailing      // zeroes      while (N % 10 == 0)     {         N = N / 10;         C += 1;     }            // Check if remaining      // N is a power of 2     if(Math.pow(2, (long)(Math.log(N) /                           (Math.log(2)))) == N)      {         D = (long)(Math.log(N) / (Math.log(2)));                    // To check the condition          // to prlong YES or NO          if (C >= D)             System.out.print("YES");         else             System.out.print("NO");      }      else         System.out.print("NO");  }    // Driver code public static void main(String args[])  {      long N = 2000000000000L;     Is_possible(N); } }    // This code is contributed by Stream_Cipher

Python

 # Python Program to check if N  # can be obtained from 1 by # repetitive multiplication  # by 10 or 20    import math    # Function to check if N can # be obtained or not def Is_possible(N):        C = 0     D = 0        # Count and remove trailing     # zeroes     while ( N % 10 == 0):         N = N / 10         C += 1        # Check if remaining     # N is a power of 2     if ( math.log(N, 2)      - int(math.log(N, 2)) == 0):            D = int(math.log(N, 2))            # To check the condition         # to print YES or NO         if (C >= D):             print("YES")                        else:             print("NO")            else:         print("NO")                # Driver Program N = 2000000000000 Is_possible(N)

C#

 // C# program to check if N  // can be obtained from 1 by  // repetitive multiplication  // by 10 or 20  using System;     class GFG{        static void Is_possible(long N) {     long C = 0;     long D = 0;            // Count and remove trailing      // zeroes      while (N % 10 == 0)     {         N = N / 10;         C += 1;     }            // Check if remaining      // N is a power of 2     if(Math.Pow(2, (long)(Math.Log(N) /                           (Math.Log(2)))) == N)      {         D = (long)(Math.Log(N) / (Math.Log(2)));                    // To check the condition          // to prlong YES or NO          if (C >= D)             Console.WriteLine("YES");         else             Console.WriteLine("NO");      }      else         Console.WriteLine("NO");  }    // Driver Code  public static void Main()  {      long N = 2000000000000L;            Is_possible(N); } }    // This code is contributed by Stream_Cipher

Output:

YES

Time Complexity: O(log10(N))
Auxiliary Space: O(1)

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Improved By : Stream_Cipher, nidhi_biet

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