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Reduce N to 1 in minimum moves by either multiplying by 2 or dividing by 6

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  • Difficulty Level : Easy
  • Last Updated : 20 Dec, 2021

Given an integer N, find the minimum number of operations to reduce N to 1 by using the following two operations:

  • Multiply N by 2
  • Divide N by 6, if N is divisible by 6

If N cannot be reduced to 1, print -1.

Examples:

Input: N = 54
Output: 5
Explanation: Perform the following operations

  • Divide N by 6 and get n = 9
  • Multiply N by 2 and get n = 18
  • Divide N by 6 and get n = 3
  • Multiply N by 2 and get n = 6
  • Divide N by 6 to get n = 1

Hence, minimum 5 operations needs to be performed to reduce 54 to 1

Input: N = 12
Output: -1

 

Approach: The task can be solved using following observations. 

  • If the number consists of primes other than 2 and 3 then the answer is -1 since N cannot be reduced to 1 with the above operations.
  • Otherwise, let count2 be the number of two’s in the factorization of n and count3 be the number of three’s in the factorization of n.
    • If count2 > count3 then the answer is -1 because we can’t get rid of all twos.
    • Otherwise, the answer is (count3−count2) + count3.

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
// of moves to reduce N to 1
void minOperations(int N)
{
    int count2 = 0, count3 = 0;
 
    // Number of 2's in the
    // factorization of N
    while (N % 2 == 0) {
        N = N / 2;
        count2++;
    }
 
    // Number of 3's in the
    // factorization of n
    while (N % 3 == 0) {
        N = N / 3;
        count3++;
    }
 
    if (N == 1 && (count2 <= count3)) {
        cout << (2 * count3) - count2;
    }
 
    // If number of 2's are greater
    // than number of 3's or
    // prime factorization of N contains
    // primes other than 2 or 3
    else {
        cout << -1;
    }
}
 
// Driver Code
int main()
{
    int N = 54;
    minOperations(N);
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the minimum number
// of moves to reduce N to 1
static void minOperations(int N)
{
    int count2 = 0, count3 = 0;
 
    // Number of 2's in the
    // factorization of N
    while (N % 2 == 0) {
        N = N / 2;
        count2++;
    }
 
    // Number of 3's in the
    // factorization of n
    while (N % 3 == 0) {
        N = N / 3;
        count3++;
    }
 
    if (N == 1 && (count2 <= count3)) {
        System.out.print((2 * count3) - count2);
    }
 
    // If number of 2's are greater
    // than number of 3's or
    // prime factorization of N contains
    // primes other than 2 or 3
    else {
        System.out.print(-1);
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 54;
    minOperations(N);
}
}
 
// This code is contributed by shikhasingrajput

Python3




# python program for the above approach
 
# Function to find the minimum number
# of moves to reduce N to 1
def minOperations(N):
 
    count2 = 0
    count3 = 0
 
    # Number of 2's in the
    # factorization of N
    while (N % 2 == 0):
        N = N // 2
        count2 += 1
 
        # Number of 3's in the
        # factorization of n
    while (N % 3 == 0):
        N = N // 3
        count3 += 1
 
    if (N == 1 and (count2 <= count3)):
        print((2 * count3) - count2)
 
        # If number of 2's are greater
        # than number of 3's or
        # prime factorization of N contains
        # primes other than 2 or 3
    else:
        print(-1)
 
# Driver Code
if __name__ == "__main__":
 
    N = 54
    minOperations(N)
 
    # This code is contributed by rakeshsahni

C#




// C# program for the above approach
using System;
class GFG {
 
    // Function to find the minimum number
    // of moves to reduce N to 1
    static void minOperations(int N)
    {
        int count2 = 0, count3 = 0;
 
        // Number of 2's in the
        // factorization of N
        while (N % 2 == 0) {
            N = N / 2;
            count2++;
        }
 
        // Number of 3's in the
        // factorization of n
        while (N % 3 == 0) {
            N = N / 3;
            count3++;
        }
 
        if (N == 1 && (count2 <= count3)) {
            Console.WriteLine((2 * count3) - count2);
        }
 
        // If number of 2's are greater
        // than number of 3's or
        // prime factorization of N contains
        // primes other than 2 or 3
        else {
            Console.WriteLine(-1);
        }
    }
 
    // Driver Code
    public static void Main()
    {
        int N = 54;
        minOperations(N);
    }
}
 
// This code is contributed by ukasp.

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to find the minimum number
// of moves to reduce N to 1
function minOperations(N)
{
    let count2 = 0, count3 = 0;
 
    // Number of 2's in the
    // factorization of N
    while (N % 2 == 0)
    {
        N = Math.floor(N / 2);
        count2++;
    }
 
    // Number of 3's in the
    // factorization of n
    while (N % 3 == 0)
    {
        N = Math.floor(N / 3);
        count3++;
    }
 
    if (N == 1 && (count2 <= count3))
    {
        document.write((2 * count3) - count2);
    }
 
    // If number of 2's are greater
    // than number of 3's or prime
    // factorization of N contains
    // primes other than 2 or 3
    else
    {
        document.write(-1);
    }
}
 
// Driver Code
let N = 54;
 
minOperations(N);
 
// This code is contributed by Potta Lokesh
 
</script>
Output
5

Time Complexity: O(logN)
Auxiliary Space: O(1)

 


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