Minimum divisions to reduce N to 1 by following given conditions

Given an integer N, the task is to find the minimum number of divisions required to reduce the number to 1 when the divisions follow the given criteria:

• Pick two integers X and Y such that X+Y  is even.
• Replace N with N/X^Y where X^Y is a divisor of N

Note: If it is impossible to reduce N to 1, then return -1.

Examples:

Input: N = 35
Output: 1
Explanation: Select X = 35 and Y = 1. X+Y = 36 which is even,  
and X^Y = 35 is a divisor of N = 35 so this is a valid choice.

Input: 44
Output: 2

Approach: The problem can be solved based on the following mathematical observation:

• If N is odd then X can be chosen as X = N and Y = 1. So X^Y = X = N and also X + Y = X + 1 = even.
• If N is even then N can be written as N = 2^p * X where p is any integer and X is an odd number (can be 1). 
  • If p is odd then it is never possible to reduce the number to 1 as 2 + p will always be odd.
  • Otherwise, it will take 2 moves to reduce the number to 1: One step to reduce the number to X and another step to reduce it from X to 1.
  • If X = 1 then no 2nd step is needed and 1 step will be sufficient.

Follow the steps mentioned below to solve the problem:

• Find if the number is odd or even.
• If the number is even then:
  • If 2 is raised to an odd power, then the answer is not possible.
  • Otherwise, get the minimum number of steps as shown in the above observation.
• If the number is odd, it takes only one step.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to find
// the minimum number of moves
int minStep(long N)
{
    if (N == 1)
        return 0;

    // If number input by user is odd
    else if (N % 2 == 1)
        return 1;
    else {
        float X;
        int count = 0;

        // Finding square root of
        // integer input by user
        X = sqrt(N);

        // Is perfect square
        if (X == round(X)) {
            return 1;
        }
        else {

            // Checking number is divisible by 2
            while (N % 2 == 0) {
                N /= 2;
                count++;
            }

            // Check value of count is
            // odd or even
            if (count % 2 == 0)
                return 2;
            else
                return -1;
        }
    }
}

// Driver code
int main()
{
    long N = 35;

    // Function call
    cout << (minStep(N));

    return 0;
}

    // This code is contributed by rakeshsahni

C

// C code to implement the approach
#include <math.h>
#include <stdio.h>

// Function to find
// the minimum number of moves
int minStep(long N)
{
    if (N == 1)
        return 0;

    // If number input by user is odd
    else if (N % 2 == 1)
        return 1;
    else {
        float X;
        int count = 0;

        // Finding square root of
        // integer input by user
        X = sqrt(N);

        // Is perfect square
        if (X == round(X)) {
            return 1;
        }
        else {

            // Checking number is divisible by 2
            while (N % 2 == 0) {
                N /= 2;
                count++;
            }
            // Check value of count is
            // odd or even
            if (count % 2 == 0)
                return 2;
            else
                return -1;
        }
    }
}

// Driver code
int main()
{
    long N = 35;

    // Function call
    printf("%d",minStep(N));
    return 0;
}

// This code is contributed by Aarohi Rai

Java

// java code to implement the approach

import java.util.*;

class GFG {

    // Function to find
    // the minimum number of moves
    public static int minStep(long N)
    {
        if (N == 1)
            return 0;

        // If number input by user is odd
        else if (N % 2 == 1)
            return 1;
        else {
            Double X;
            int count = 0;

            // Finding square root of
            // integer input by user
            X = Math.sqrt(N);

            // Is perfect square
            if (X == Math.round(X)) {
                return 1;
            }
            else {

                // Checking number is divisible by 2
                while (N % 2 == 0) {
                    N /= 2;
                    count++;
                }

                // Check value of count is
                // odd or even
                if (count % 2 == 0)
                    return 2;
                else
                    return -1;
            }
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        long N = 35;

        // Function call
        System.out.println(minStep(N));
    }
}

Python3

# Python code to implement the approach
import math

# Function to find
# the minimum number of moves
def minStep(N):
    if N == 1:
        return 0
    # If number input by user is odd
    elif N % 2 == 1:
        return 1
    else:
        X = 0.0
        count = 0
        # Finding square root of
        # integer input by user
        X = math.sqrt(N)
        # Is perfect square
        if X == round(X):
            return 1

        else:
            # Checking number is divisible by 2
            while N % 2 == 0:
                N /= 2
                count += 1

            # Check value of count is
            # odd or even
            if count % 2 == 0:
                return 2
            else:
                return -1


# Driver code
if __name__ == "__main__":
    N = 35
    # Function call
    print(minStep(N))


# This code is contributed by Rohit Pradhan

C#

// C# code to implement the approach
using System;

class GFG {

    // Function to find
    // the minimum number of moves
    static int minStep(long N)
    {
        if (N == 1)
            return 0;

        // If number input by user is odd
        else if (N % 2 == 1)
            return 1;
        else {
            double X = 0;
            int count = 0;

            // Finding square root of
            // integer input by user
            X = Math.Sqrt(N);

            // Is perfect square
            if (X == Math.Round(X)) {
                return 1;
            }
            else {

                // Checking number is divisible by 2
                while (N % 2 == 0) {
                    N /= 2;
                    count++;
                }

                // Check value of count is
                // odd or even
                if (count % 2 == 0)
                    return 2;
                else
                    return -1;
            }
        }
    }

    // Driver code
    public static void Main()
    {
        long N = 35;

        // Function call
        Console.WriteLine(minStep(N));
    }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>

// JavaScript code to implement the approach

// Function to find
// the minimum number of moves
function minStep(N)
{
    if (N == 1)
        return 0;

    // If number input by user is odd
    else if (N % 2 == 1)
        return 1;
    else {
        let X;
        let count = 0;

        // Finding square root of
        // integer input by user
        X = Math.sqrt(N);

        // Is perfect square
        if (X == Math.round(X)) {
            return 1;
        }
        else {

            // Checking number is divisible by 2
            while (N % 2 == 0) {
                N = Math.floor(N/2);
                count++;
            }

            // Check value of count is
            // odd or even
            if (count % 2 == 0)
                return 2;
            else
                return -1;
        }
    }
}

// Driver code
let N = 35;

// Function call
document.write(minStep(N),"</br>");

// This code is contributed by shinjanpatra

</script>

Output

1

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