# Minimize divisions by 2, 3, or 5 required to make two given integers equal

• Last Updated : 02 Nov, 2021

Given two integers X and Y, the task is to make X and Y equal by dividing either of X or Y by 2, 3, or 5, minimum number of times, if found to be divisible. If the two integers can be made equal, then print “-1”.

Examples:

Input: X = 15, Y = 20
Output: 3
Explanation:
Operation 1: Reduce X(= 15) to 15/3 i.e., X = 15/3 = 5.
Operation 2: Reduce Y(= 20) to 20/2 i.e., Y = 20/2 = 10.
Operation 3: Reduce Y(= 20) to 10/2 i.e., Y = 10/2 = 5.
After the above operations, X and Y are equal i.e., 5.
Therefore, the count of operation required is 3.

Input: X = 6, Y = 6
Output: 0

Approach: The given problem can be solved greedily. The idea is to reduce the given integers to their GCD in order to make them equal. Follow the steps below to solve the problem:

• Find the Greatest Common Divisor (GCD) of X and Y and divide X and Y by their GCD.
• Initialize a variable, say count, to store the number of divisions required.
• Iterate until X is not equal to Y and perform the following steps:
• If the value of X is greater than Y, then swap X and Y.
• If the larger number(i.e., Y) is divisible by 2, 3, or 5, then divide it by that number. Increment count by 1. Otherwise, if it is not possible to make both the numbers equal, print “-1” and break out of the loop.
• After completing the above steps, if both the numbers can be made equal, then print the value of count as the minimum number of divisions required to make X and Y equal.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to calculate``// GCD of two numbers``int` `gcd(``int` `a, ``int` `b)``{``    ``// Base Case``    ``if` `(b == 0) {``        ``return` `a;``    ``}` `    ``// Calculate GCD recursively``    ``return` `gcd(b, a % b);``}` `// Function to count the minimum``// number of divisions required``// to make X and Y equal``void` `minimumOperations(``int` `X, ``int` `Y)``{``    ``// Calculate GCD of X and Y``    ``int` `GCD = gcd(X, Y);` `    ``// Divide X and Y by their GCD``    ``X = X / GCD;``    ``Y = Y / GCD;` `    ``// Stores the number of divisions``    ``int` `count = 0;` `    ``// Iterate until X != Y``    ``while` `(X != Y) {` `        ``// Maintain the order X <= Y``        ``if` `(Y > X) {``            ``swap(X, Y);``        ``}` `        ``// If X is divisible by 2,``        ``// then divide X by 2``        ``if` `(X % 2 == 0) {``            ``X = X / 2;``        ``}` `        ``// If X is divisible by 3,``        ``// then divide X by 3``        ``else` `if` `(X % 3 == 0) {``            ``X = X / 3;``        ``}` `        ``// If X is divisible by 5,``        ``// then divide X by 5``        ``else` `if` `(X % 5 == 0) {``            ``X = X / 5;``        ``}` `        ``// If X is not divisible by``        ``// 2, 3, or 5, then print -1``        ``else` `{``            ``cout << ``"-1"``;``            ``return``;``        ``}` `        ``// Increment count by 1``        ``count++;``    ``}` `    ``// Print the value of count as the``    ``// minimum number of operations``    ``cout << count;``}` `// Driver Code``int` `main()``{``    ``int` `X = 15, Y = 20;``    ``minimumOperations(X, Y);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG{` `// Function to calculate``// GCD of two numbers``static` `int` `gcd(``int` `a, ``int` `b)``{``    ` `    ``// Base Case``    ``if` `(b == ``0``)``    ``{``        ``return` `a;``    ``}` `    ``// Calculate GCD recursively``    ``return` `gcd(b, a % b);``}` `// Function to count the minimum``// number of divisions required``// to make X and Y equal``static` `void` `minimumOperations(``int` `X, ``int` `Y)``{``    ` `    ``// Calculate GCD of X and Y``    ``int` `GCD = gcd(X, Y);` `    ``// Divide X and Y by their GCD``    ``X = X / GCD;``    ``Y = Y / GCD;` `    ``// Stores the number of divisions``    ``int` `count = ``0``;` `    ``// Iterate until X != Y``    ``while` `(X != Y)``    ``{``        ` `        ``// Maintain the order X <= Y``        ``if` `(Y > X)``        ``{``            ``int` `t = X;``            ``X = Y;``            ``Y = t;``        ``}` `        ``// If X is divisible by 2,``        ``// then divide X by 2``        ``if` `(X % ``2` `== ``0``)``        ``{``            ``X = X / ``2``;``        ``}` `        ``// If X is divisible by 3,``        ``// then divide X by 3``        ``else` `if` `(X % ``3` `== ``0``)``        ``{``            ``X = X / ``3``;``        ``}` `        ``// If X is divisible by 5,``        ``// then divide X by 5``        ``else` `if` `(X % ``5` `== ``0``)``        ``{``            ``X = X / ``5``;``        ``}` `        ``// If X is not divisible by``        ``// 2, 3, or 5, then print -1``        ``else``        ``{``            ``System.out.print(``"-1"``);``            ``return``;``        ``}` `        ``// Increment count by 1``        ``count += ``1``;``    ``}` `    ``// Print the value of count as the``    ``// minimum number of operations``    ``System.out.println(count);``}` `// Driver Code``static` `public` `void` `main(String args[])``{``    ``int` `X = ``15``, Y = ``20``;``    ` `    ``minimumOperations(X, Y);``}``}` `// This code is contributed by ipg2016107`

## Python3

 `# Python3 program for the above approach` `# Function to calculate``# GCD of two numbers``def` `gcd(a, b):``    ` `    ``# Base Case``    ``if` `(b ``=``=` `0``):``        ``return` `a` `    ``# Calculate GCD recursively``    ``return` `gcd(b, a ``%` `b)` `# Function to count the minimum``# number of divisions required``# to make X and Y equal``def` `minimumOperations(X, Y):``    ` `    ``# Calculate GCD of X and Y``    ``GCD ``=` `gcd(X, Y)` `    ``# Divide X and Y by their GCD``    ``X ``=` `X ``/``/` `GCD``    ``Y ``=` `Y ``/``/` `GCD` `    ``# Stores the number of divisions``    ``count ``=` `0` `    ``# Iterate until X != Y``    ``while` `(X !``=` `Y):` `        ``# Maintain the order X <= Y``        ``if` `(Y > X):``            ``X, Y ``=` `Y, X` `        ``# If X is divisible by 2,``        ``# then divide X by 2``        ``if` `(X ``%` `2` `=``=` `0``):``            ``X ``=` `X ``/``/` `2` `        ``# If X is divisible by 3,``        ``# then divide X by 3``        ``elif` `(X ``%` `3` `=``=` `0``):``            ``X ``=` `X ``/``/` `3` `        ``# If X is divisible by 5,``        ``# then divide X by 5``        ``elif` `(X ``%` `5` `=``=` `0``):``            ``X ``=` `X ``/``/` `5``            ` `        ``# If X is not divisible by``        ``# 2, 3, or 5, then pr-1``        ``else``:``            ``print``(``"-1"``)``            ``return` `        ``# Increment count by 1``        ``count ``+``=` `1` `    ``# Print the value of count as the``    ``# minimum number of operations``    ``print` `(count)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``X, Y ``=` `15``, ``20``    ` `    ``minimumOperations(X, Y)` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach``using` `System;` `public` `class` `GFG``{` `// Function to calculate``// GCD of two numbers``static` `int` `gcd(``int` `a, ``int` `b)``{``  ` `    ``// Base Case``    ``if` `(b == 0) {``        ``return` `a;``    ``}` `    ``// Calculate GCD recursively``    ``return` `gcd(b, a % b);``}` `// Function to count the minimum``// number of divisions required``// to make X and Y equal``static` `void` `minimumOperations(``int` `X, ``int` `Y)``{``  ` `    ``// Calculate GCD of X and Y``    ``int` `GCD = gcd(X, Y);` `    ``// Divide X and Y by their GCD``    ``X = X / GCD;``    ``Y = Y / GCD;` `    ``// Stores the number of divisions``    ``int` `count = 0;` `    ``// Iterate until X != Y``    ``while` `(X != Y) {` `        ``// Maintain the order X <= Y``        ``if` `(Y > X) {``            ``int` `t = X;``            ``X = Y;``            ``Y = t;``        ``}` `        ``// If X is divisible by 2,``        ``// then divide X by 2``        ``if` `(X % 2 == 0) {``            ``X = X / 2;``        ``}` `        ``// If X is divisible by 3,``        ``// then divide X by 3``        ``else` `if` `(X % 3 == 0) {``            ``X = X / 3;``        ``}` `        ``// If X is divisible by 5,``        ``// then divide X by 5``        ``else` `if` `(X % 5 == 0) {``            ``X = X / 5;``        ``}` `        ``// If X is not divisible by``        ``// 2, 3, or 5, then print -1``        ``else` `{``            ``Console.WriteLine(``"-1"``);``            ``return``;``        ``}` `        ``// Increment count by 1``        ``count++;``    ``}` `    ``// Print the value of count as the``    ``// minimum number of operations``    ``Console.WriteLine(count);``}` `// Driver Code``static` `public` `void` `Main()``{``    ``int` `X = 15, Y = 20;``    ``minimumOperations(X, Y);``}``}` `// This code is contributed by sanjoy_62.`

## Javascript

 ``
Output:
`3`

Time Complexity: O(log(max(X, Y)))
Auxiliary Space: O(1)

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