Given A numbers of white and B numbers of black balls. You need to have X number of white and Y number of black balls (A <= X, B <= Y) to win the game by doing some(zero or more) operations.
In one operation: At any moment if you have p white and q black balls then at that moment you can buy q white or p black balls.
Find if it’s possible to have X white and Y black balls at the end or not.
Examples:
Input: A = 1, B = 1, X = 3, Y = 8 Output: POSSIBLE Explanation: The steps are, (1, 1)->(1, 2)->(3, 2)->(3, 5)->(3, 8) Input: A = 3, Y = 2, X = 4, Y = 6 Output: NOT POSSIBLE
Approach:
We have to solve this problem using the property of gcd. Let’s see how.
- Initially, we have A white and B black ball. We have to get rest of the X-A Red balls and Y-B Black balls.
- Below is the property of gcd of two numbers that we will use,
gcd(x, y) = gcd(x + y, y) gcd(x, y) = gcd(x, y + x)
- This property is the same as the operation which is mentioned in the question. So from here, we get that if the gcd of the final state is the same as gcd of initial state then it is always possible to reach the goal, otherwise not.
Below is the implementation of the above approach:
C++
// C++ program to Find is it possible// to have X white and Y black// balls at the end.#include <bits/stdc++.h>using namespace std;// Recursive function to return // gcd of a and bint gcd(int a, int b){ if (b == 0) return a; return gcd(b, a % b);}// Function returns if it's // possible to have X white // and Y black balls or not.void IsPossible(int a, int b, int x, int y){ // Finding gcd of (x, y) // and (a, b) int final = gcd(x, y); int initial = gcd(a, b); // If gcd is same, it's always // possible to reach (x, y) if (initial == final) { cout << "POSSIBLE\n"; } else { // Here it's never possible // if gcd is not same cout << "NOT POSSIBLE\n"; }}// Driver Codeint main(){ int A = 1, B = 2, X = 4, Y = 11; IsPossible(A, B, X, Y); A = 2, B = 2, X = 3, Y = 6; IsPossible(A, B, X, Y); return 0;} |
Java
// Java program to Find is it possible// to have X white and Y black// balls at the end.import java.io.*;class GFG{// Recursive function to return // gcd of a and bstatic int gcd(int a, int b){ if (b == 0) return a; return gcd(b, a % b);}// Function returns if it's // possible to have X white // and Y black balls or not.static void IsPossible(int a, int b, int x, int y){ // Finding gcd of (x, y) // and (a, b) int g = gcd(x, y); int initial = gcd(a, b); // If gcd is same, it's always // possible to reach (x, y) if (initial == g) { System.out.print("POSSIBLE\n"); } else { // Here it's never possible // if gcd is not same System.out.print("NOT POSSIBLE\n"); }}// Driver codepublic static void main(String args[]){ int A = 1, B = 2, X = 4, Y = 11; IsPossible(A, B, X, Y); A = 2; B = 2; X = 3; Y = 6; IsPossible(A, B, X, Y);}}// This code is contributed by shivanisinghss2110 |
Python3
# Python3 program to find is it possible # to have X white and Y black # balls at the end. # Recursive function to return # gcd of a and b def gcd(a, b) : if (b == 0) : return a; return gcd(b, a % b); # Function returns if it's # possible to have X white # and Y black balls or not. def IsPossible(a, b, x, y) : # Finding gcd of (x, y) # and (a, b) final = gcd(x, y); initial = gcd(a, b); # If gcd is same, it's always # possible to reach (x, y) if (initial == final) : print("POSSIBLE"); else : # Here it's never possible # if gcd is not same print("NOT POSSIBLE"); # Driver Code if __name__ == "__main__" : A = 1; B = 2; X = 4; Y = 11; IsPossible(A, B, X, Y); A = 2; B = 2; X = 3; Y = 6; IsPossible(A, B, X, Y); # This code is contributed by AnkitRai01 |
C#
// C# program to Find is it possible// to have X white and Y black// balls at the end.using System; using System.Linq;class GFG { // Recursive function to return // gcd of a and bstatic int gcd(int a, int b){ if (b == 0) return a; return gcd(b, a % b);}// Function returns if it's // possible to have X white // and Y black balls or not.static void IsPossible(int a, int b, int x, int y){ // Finding gcd of (x, y) // and (a, b) int g = gcd(x, y); int initial = gcd(a, b); // If gcd is same, it's always // possible to reach (x, y) if (initial == g) { Console.Write("POSSIBLE\n"); } else { // Here it's never possible // if gcd is not same Console.Write("NOT POSSIBLE\n"); }}// Driver codestatic public void Main(){ int A = 1, B = 2; int X = 4, Y = 11; IsPossible(A, B, X, Y); A = 2; B = 2; X = 3; Y = 6; IsPossible(A, B, X, Y);}}// This code is contributed by shivanisinghss2110 |
Javascript
<script> // Javascript program to Find is it possible // to have X white and Y black // balls at the end. // Recursive function to return // gcd of a and b function gcd(a, b) { if (b == 0) return a; return gcd(b, a % b); } // Function returns if it's // possible to have X white // and Y black balls or not. function IsPossible(a, b, x, y) { // Finding gcd of (x, y) // and (a, b) let final = gcd(x, y); let initial = gcd(a, b); // If gcd is same, it's always // possible to reach (x, y) if (initial == final) { document.write("POSSIBLE" + "</br>"); } else { // Here it's never possible // if gcd is not same document.write("NOT POSSIBLE" + "</br>"); } } let A = 1, B = 2, X = 4, Y = 11; IsPossible(A, B, X, Y); A = 2, B = 2, X = 3, Y = 6; IsPossible(A, B, X, Y); // This code is contributed by divyesh072019.</script> |
Output:
POSSIBLE NOT POSSIBLE
Time Complexity: O(log(max(A, B, X, Y)))
Auxiliary Space: O(1)
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