Given two positive integers X and Y, the task is to check if X can be converted to Y or not by repeatedly changing the value of X to (3 * X / 2) (if X is even) or (X – 1). If it is possible to convert X into Y, then print “Yes”. Otherwise, print “No”.
Examples:
Input: X = 6, Y = 8
Output: Yes
Explanation:
Operation 1: Convert X(= 6) to 3*X/2 ( = 3 * (6 / 2) = 9).
Operation 2: Convert X(= 9) to (X – 1) (= 8).
Therefore, the total number of operations required is 2.Input: X = 3, Y = 6
Output: No
Approach: The given problem can be solved based on the following observations:
- If the value of X is at least Y, then X can always be converted to Y using the second operation, i.e. reducing X by 1.
- If X is even, then it can be converted to (3 * (X / 2)), which is greater than X for all even numbers greater than 0.
- If the value of X is odd, then X can be converted to (X – 1), which can be converted into (3 * (X – 1)/2), which is greater than X.
- Therefore, from the above observations, the conversion of X into Y is always possible for X > 3.
- Following base cases are required to be considered:
- X = 1: Conversion possible only if Y = 1.
- X = 2 or X = 3: Conversion possible only if Y ? 3.
- In all the other cases, conversion is not possible.
Follow the steps below to solve the problem:
- If the value X is greater than 4, then print “Yes”.
- If the value X is 1 and Y is 1, then print “Yes”.
- If the value X is 2 or 3 and Y is less than 4, then print “Yes”.
- Otherwise, print “No” for all the other cases.
Below is the implementation of above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if X can be// made equal to Y by converting// X to (3*X/2) or (X - 1)void check(int X, int Y){ // Conditions for possible conversion if (X > 3) { cout << "Yes"; } else if (X == 1 and Y == 1) { cout << "Yes"; } else if (X == 2 and Y <= 3) { cout << "Yes"; } else if (X == 3 and Y <= 3) { cout << "Yes"; } // Otherwise, conversion // is not possible else { cout << "No"; }}// Driver Codeint main(){ int X = 6, Y = 8; check(X, Y); return 0;} |
Java
// Java program for the above approachclass GFG{ // Function to check if X can be// made equal to Y by converting// X to (3*X/2) or (X - 1)static void check(int X, int Y){ // Conditions for possible conversion if (X > 3) { System.out.print("Yes"); } else if (X == 1 && Y == 1) { System.out.print("Yes"); } else if (X == 2 && Y <= 3) { System.out.print("Yes"); } else if (X == 3 && Y <= 3) { System.out.print("Yes"); } // Otherwise, conversion // is not possible else { System.out.print("No"); }}// Driver Codepublic static void main (String[] args) { int X = 6, Y = 8; check(X, Y);}}// This code is contributed by AnkThon |
Python3
# Python3 program for the above approach# Function to check if X can be# made equal to Y by converting# X to (3*X/2) or (X - 1)def check(X, Y): # Conditions for possible conversion if (X > 3): print("Yes") elif (X == 1 and Y == 1): print("Yes") elif (X == 2 and Y <= 3): print("Yes") elif (X == 3 and Y <= 3): print("Yes") # Otherwise, conversion # is not possible else: print("No")# Driver Codeif __name__ == '__main__': X = 6 Y = 8 check(X, Y)# This code is contributed by ipg2016107 |
C#
// C# program for the above approachusing System;class GFG{// Function to check if X can be// made equal to Y by converting// X to (3*X/2) or (X - 1)static void check(int X, int Y){ // Conditions for possible conversion if (X > 3) { Console.WriteLine("Yes"); } else if (X == 1 && Y == 1) { Console.WriteLine("Yes"); } else if (X == 2 && Y <= 3) { Console.WriteLine("Yes"); } else if (X == 3 && Y <= 3) { Console.WriteLine("Yes"); } // Otherwise, conversion // is not possible else { Console.WriteLine("No"); }}// Driver Codepublic static void Main(string[] args){ int X = 6, Y = 8; check(X, Y);}}// This code is contributed by avijitmondal1998 |
Javascript
<script> // JavaScript program for the above approach // Function to check if X can be // made equal to Y by converting // X to (3*X/2) or (X - 1) function check(X, Y) { // Conditions for possible conversion if (X > 3) { document.write("Yes"); } else if (X == 1 && Y == 1) { document.write("Yes"); } else if (X == 2 && Y <= 3) { document.write("Yes"); } else if (X == 3 && Y <= 3) { document.write("Yes"); } // Otherwise, conversion // is not possible else { document.write("No"); } } let X = 6, Y = 8; check(X, Y);</script> |
Output:
Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
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