Check if X and Y can be equal in N steps by dividing them with their factors

Given positive integers, X, Y, and N, the task is to check whether X can be made equal to Y in exactly N operations wherein each operation:

• X can be divided by any of its factors other than 1.
• Y can be divided by any of its factors other than 1.

Examples:

Input: X = 4, Y = 21, N = 3
Output: Yes
Explanation: Both X and Y can be made equal by doing the following operations: 
X = X/4, Y = Y/3, Y = Y/7 which results in both X and Y as 1. 
There are several other ways also to make both equal in three operations.
Like X = X/2, X = X/2, Y = Y/21.

Input: X = 5, Y = 17, N = 3
Output: No

Approach: The idea to solve the problem is based on the following observation:

• The problem can be solved by making both the numbers 1. 
• Minimum number of times to divide a number to get 1 is to divide the number by itself. And the maximum number of times to get 1 is the sum of exponents of all prime factors.
• Any natural number N can be written in terms of the product of its prime factors as:
  N = p₁ⁿ¹.p₂ⁿ². . . p_k^nk, where p_1, p₂ . . . p_k are distinct prime numbers.
• So the maximum number of times N can be divided is equal to n1 + n2 + …. + nk.
• If the maximum divisor of both X and Y combined is greater or equal to N, both can be made equal otherwise not.

Follow the steps mentioned below to implement the above observation:

• If N is equal to 1:
  • If one among X and Y is the factor of the other, then they can be made equal.
  • Otherwise, they cannot be made equal.
• Otherwise, if N is less than or equal to the maximum number of factors of X and Y then they can be made equal.
• Otherwise, no solution is possible.

Below is the implementation of the above approach in an optimized way:

C++

// C++ code to implement above approach

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

// Function to find total
// number of times we can
// divide a number before making it 1.
int count(int val)
{
    int ans = 0;

    // Checking how many times
    // the number can be
    // divided by 2.
    while (val % 2 == 0) {
        ans++;
        val = val / 2;
    }

    // Iterating through each number from
    // 3 to sqrt(val) and finding how many
    // times the number can be divide
    for (int i = 3; i <= sqrt(val); i = i + 2) {
        while (val % i == 0) {
            ans++;
            val = val / i;
        }
    }
    if (val > 2)
        ans++;

    return ans;
}

// Function to return if
// we can make x and y equal or not
int solve(int x, int y, int n)
{
    if (n == 0) {
        if (x == y) {
            cout << "Yes";
        }
        else {
            cout << "No";
        }
    }

    else if (n == 1) {
        if ((x % y == 0 || y % x == 0)
            && x != y) {
            cout << "Yes";
        }
        else {
            cout << "No";
        }
    }

    else {
        if (count(x) + count(y) >= n) {
            cout << "Yes";
        }
        else {
            cout << "No";
        }
    }
}

// Driver code
int main()
{
    int X = 4, Y = 21, N = 3;

    // Function call
    solve(X, Y, N);
    return 0;
}

Java

// Java code to implement above approach
import java.io.*;

class GFG
{

  // Function to find total
  // number of times we can
  // divide a number before making it 1.
  static int count(int val)
  {
    int ans = 0;

    // Checking how many times
    // the number can be
    // divided by 2.
    while (val % 2 == 0) {
      ans++;
      val = val / 2;
    }

    // Iterating through each number from
    // 3 to sqrt(val) and finding how many
    // times the number can be divide
    for (int i = 3; i <= Math.sqrt(val); i = i + 2) {
      while (val % i == 0) {
        ans++;
        val = val / i;
      }
    }
    if (val > 2)
      ans++;

    return ans;
  }

  // Function to return if
  // we can make x and y equal or not
  static void solve(int x, int y, int n)
  {
    if (n == 0) {
      if (x == y) {
        System.out.print("Yes");
      }
      else {
        System.out.print("No");
      }
    }

    else if (n == 1) {
      if ((x % y == 0 || y % x == 0)
          && x != y) {
        System.out.print("Yes");
      }
      else {
        System.out.print("No");
      }
    }

    else {
      if (count(x) + count(y) >= n) {
        System.out.print("Yes");
      }
      else {
        System.out.print("No");
      }
    }
  }

  // Driver code
  public static void main (String[] args) {
    int X = 4, Y = 21, N = 3;

    // Function call
    solve(X, Y, N);
  }
}

// This code is contributed by hrithikgarg03188.

Python3

# Python3 code to implement above approach

# Function to find total
# number of times we can
# divide a number before making it 1.
def count(val):

    ans = 0

    # Checking how many times
    # the number can be
    # divided by 2.
    while (val % 2 == 0):
        ans += 1
        val = val // 2

    # Iterating through each number from
    # 3 to sqrt(val) and finding how many
    # times the number can be divide
    for i in range(3, int(val**0.5)+1, 2):
        while (val % i == 0):
            ans += 1
            val = val // i

    if (val > 2):
        ans += 1

    return ans


# Function to return if
# we can make x and y equal or not
def solve(x,  y,  n):

    if (n == 0):
        if (x == y):
            print("Yes")

        else:
            print("No")

    elif (n == 1):
        if ((x % y == 0 or y % x == 0) and x != y):
            print("Yes")
        else:
            print("No")
    else:
        if (count(x) + count(y) >= n):
            print("Yes")
        else:
            print("No")

# Driver code
if __name__ == "__main__":
    x = 4
    y = 21
    n = 3
    solve(x, y, n)

    # This code is contributed by amnindersingh1414.

C#

// C# code to implement above approach
using System;
public class GFG {

  // Function to find total
  // number of times we can
  // divide a number before making it 1.
  static int count(int val)
  {
    int ans = 0;

    // Checking how many times
    // the number can be
    // divided by 2.
    while (val % 2 == 0) {
      ans++;
      val = val / 2;
    }

    // Iterating through each number from
    // 3 to sqrt(val) and finding how many
    // times the number can be divide
    for (int i = 3; i <= Math.Sqrt(val); i = i + 2) {
      while (val % i == 0) {
        ans++;
        val = val / i;
      }
    }
    if (val > 2)
      ans++;

    return ans;
  }

  // Function to return if
  // we can make x and y equal or not
  static void solve(int x, int y, int n)
  {
    if (n == 0) {
      if (x == y) {
        Console.Write("Yes");
      }
      else {
        Console.Write("No");
      }
    }

    else if (n == 1) {
      if ((x % y == 0 || y % x == 0) && x != y) {
        Console.Write("Yes");
      }
      else {
        Console.Write("No");
      }
    }

    else {
      if (count(x) + count(y) >= n) {
        Console.Write("Yes");
      }
      else {
        Console.Write("No");
      }
    }
  }

  // Driver code
  public static int Main()
  {
    int X = 4, Y = 21, N = 3;

    // Function call
    solve(X, Y, N);
    return 0;
  }
}

// This code is contributed by Taranpreet

Javascript

<script>
    // JavaScript code to implement above approach

    // Function to find total
    // number of times we can
    // divide a number before making it 1.
    const count = (val) => {
        let ans = 0;

        // Checking how many times
        // the number can be
        // divided by 2.
        while (val % 2 == 0) {
            ans++;
            val = parseInt(val / 2);
        }

        // Iterating through each number from
        // 3 to sqrt(val) and finding how many
        // times the number can be divide
        for (let i = 3; i <= parseInt(Math.sqrt(val)); i = i + 2) {
            while (val % i == 0) {
                ans++;
                val = parseInt(val / i);
            }
        }
        if (val > 2)
            ans++;

        return ans;
    }

    // Function to return if
    // we can make x and y equal or not
    const solve = (x, y, n) => {
        if (n == 0) {
            if (x == y) {
                document.write("Yes");
            }
            else {
                document.write("No");
            }
        }

        else if (n == 1) {
            if ((x % y == 0 || y % x == 0)
                && x != y) {
                document.write("Yes");
            }
            else {
                document.write("No");
            }
        }

        else {
            if (count(x) + count(y) >= n) {
                document.write("Yes");
            }
            else {
                document.write("No");
            }
        }
    }

    // Driver code
    let X = 4, Y = 21, N = 3;

    // Function call
    solve(X, Y, N);

// This code is contributed by rakeshsahni

</script>

Output

Yes

Time complexity: O(√X)+ √Y)
Auxiliary Space: O(1)