Minimum division by 10 and multiplication by 2 required to reduce given number to 1

Last Updated : 26 Apr, 2021

Given an integer N, the task is to reduce N to 1 by a minimum number of operations involving multiplication by 2 and division by 10. If 1 cannot be obtained, then print “-1”.

Examples:

Input: N = 5
Output: 2
Explanation:
Below are the operations performed:
1st operation: Multiply N by 2. Therefore, N = 5 * 2 = 10.
2nd operation: Divide N by 10. Therefore, N = 10/10 = 1.
Therefore, minimum number of operations required is 2.

Input: N = 4
Output: -1

Approach: The idea is to check prime factors of the given number M. If the given number has prime factors other than 2 and 5, then it is not possible to reduce the given number to 1 by the given operations. If the count of 2 exceeds that of 5 in its prime factors, then it is not possible to reduce N to 1 as all powers of 2 can’t be reduced.
Follow the steps below to solve the problem:

Count the number of 2s present in prime factors of N and store it in a variable, say cnt2, and update N to N / 2^cnt2.
Count the number of 5s present in prime factors of N and store it in a variable, say cnt5, and update N to N / 5^cnt5.
After completing the above steps, if N is 1 and cnt2 ≤ cnt5, then the minimum number of steps required is 2 * cnt5 – cnt2.
Otherwise, print “-1” as N can’t be reduced to 1 with the given operations.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum number
// operations required to reduce N to 1
int minimumMoves(int n)
{
 
    // Stores count of powers of 2 and 5
    int cnt2 = 0, cnt5 = 0;
 
    // Calculating the primefactors 2
    while (n % 2 == 0) {
        n /= 2;
        cnt2++;
    }
 
    // Calculating the primefactors 5
    while (n % 5 == 0) {
        n /= 5;
        cnt5++;
    }
 
    // If n is 1 and cnt2 <= cnt5
    if (n == 1 && cnt2 <= cnt5) {
 
        // Return the minimum operations
        return 2 * cnt5 - cnt2;
    }
 
    // Otherwise, n can't be reduced
    else
        return -1;
}
 
// Driver Code
int main()
{
    // Given Number N
    int N = 25;
 
    // Function Call
    cout << minimumMoves(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the minimum number
// operations required to reduce N to 1
static int minimumMoves(int n)
{
     
    // Stores count of powers of 2 and 5
    int cnt2 = 0, cnt5 = 0;
 
    // Calculating the primefactors 2
    while (n % 2 == 0)
    {
        n /= 2;
        cnt2++;
    }
 
    // Calculating the primefactors 5
    while (n % 5 == 0) 
    {
        n /= 5;
        cnt5++;
    }
 
    // If n is 1 and cnt2 <= cnt5
    if (n == 1 && cnt2 <= cnt5)
    {
         
        // Return the minimum operations
        return 2 * cnt5 - cnt2;
    }
 
    // Otherwise, n can't be reduced
    else
        return -1;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Number N
    int N = 25;
 
    // Function Call
    System.out.print(minimumMoves(N));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
 
# Function to find the minimum number
# operations required to reduce N to 1
def minimumMoves(n):
 
    # Stores count of powers of 2 and 5
    cnt2 = 0
    cnt5 = 0
 
    # Calculating the primefactors 2
    while (n % 2 == 0):
        n //= 2
        cnt2 += 1
 
    # Calculating the primefactors 5
    while (n % 5 == 0):
        n //= 5
        cnt5 += 1
 
    # If n is 1 and cnt2 <= cnt5
    if (n == 1 and cnt2 <= cnt5):
         
        # Return the minimum operations
        return 2 * cnt5 - cnt2
 
    # Otherwise, n can't be reduced
    else:
        return -1
 
# Driver Code
if __name__ == '__main__':
     
    # Given Number N
    N = 25
 
    # Function Call
    print(minimumMoves(N))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the minimum number
// operations required to reduce N to 1
static int minimumMoves(int n)
{
     
    // Stores count of powers of 2 and 5
    int cnt2 = 0, cnt5 = 0;
 
    // Calculating the primefactors 2
    while (n % 2 == 0)
    {
        n /= 2;
        cnt2++;
    }
 
    // Calculating the primefactors 5
    while (n % 5 == 0) 
    {
        n /= 5;
        cnt5++;
    }
 
    // If n is 1 and cnt2 <= cnt5
    if (n == 1 && cnt2 <= cnt5)
    {
         
        // Return the minimum operations
        return 2 * cnt5 - cnt2;
    }
 
    // Otherwise, n can't be reduced
    else
        return -1;
}
 
// Driver Code
public static void Main()
{
     
    // Given Number N
    int N = 25;
 
    // Function Call
    Console.WriteLine(minimumMoves(N));
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript

<script>
// Javascript program to implement
// the above approach
 
// Function to find the minimum number
// operations required to reduce N to 1
function minimumMoves(n)
{
      
    // Stores count of powers of 2 and 5
    let cnt2 = 0, cnt5 = 0;
  
    // Calculating the primefactors 2
    while (n % 2 == 0)
    {
        n /= 2;
        cnt2++;
    }
  
    // Calculating the primefactors 5
    while (n % 5 == 0)
    {
        n /= 5;
        cnt5++;
    }
  
    // If n is 1 and cnt2 <= cnt5
    if (n == 1 && cnt2 <= cnt5)
    {
          
        // Return the minimum operations
        return 2 * cnt5 - cnt2;
    }
  
    // Otherwise, n can't be reduced
    else
        return -1;
}
 
    // Driver Code
     
    // Given Number N
    let N = 25;
  
    // Function Call
    document.write(minimumMoves(N));
      
</script>

Output:

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

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Table of Contents

Minimum division by 10 and multiplication by 2 required to reduce given number to 1

C++

Java

Python3

C#

Javascript

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