Given a pair of coordinates (X1, Y1)(source) and (X2, Y2)(destination), the task is to check if it is possible to reach the destination form the source by the following movements from any cell (X, Y):
- (X + Y, Y)
- (X, Y + X)
Note: All coordinates are positive and can be as large as 1018.
Examples:
Input: X1 = 2, Y1 = 10, X2 = 26, Y2 = 12
Output: YesExplanation: Possible path: (2, 10) ? (2, 12) ? (14, 12) ? (26, 12)
Therefore, a path exists between source and destination.Input: X1 = 20, Y1 = 10, X2 = 6, Y2 = 12
Output: No
Naive Approach: The simplest approach to solve the problem is by using recursion. Refer to the article check if a destination is reachable from source with two movements allowed for the recursive approach.
Efficient Approach: The main idea is to check if a path from the destination coordinates (X2, Y2) to the source (X1, Y1) exists or not.
Follow the steps below to solve the problem:
- Keep subtracting the smallest of (X2, Y2) from the largest of (X2, Y2) and stop if X2 becomes less than X1 or Y2 becomes less than Y1.
- Now, compare (X1, Y1) and modified (X2, Y2). If X1 is equal to X2 and Y1 is equal to Y2, then print “Yes“.
- If X1 is not equal to X2 or Y1 is equal, not Y2, then print “No“.
To optimize the complexity of the subtraction operation, the modulus operation can be used instead. Simply perform x2 = x2 % y2 and y2 = y2 % x2 and check for the necessary condition mentioned above.
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Check if (x2, y2) can be reached// from (x1, y1)bool isReachable(long long x1, long long y1, long long x2, long long y2){ while (x2 > x1 && y2 > y1) { // Reduce x2 by y2 until it is // less than or equal to x1 if (x2 > y2) x2 %= y2; // Reduce y2 by x2 until it is // less than or equal to y1 else y2 %= x2; } // If x2 is reduced to x1 if (x2 == x1) // Check if y2 can be // reduced to y1 or not return (y2 - y1) >= 0 && (y2 - y1) % x1 == 0; // If y2 is reduced to y1 else if (y2 == y1) // Check if x2 can be // reduced to x1 or not return (x2 - x1) >= 0 && (x2 - x1) % y1 == 0; else return 0;}// Driver Codeint main(){ long long source_x = 2, source_y = 10; long long dest_x = 26, dest_y = 12; if (isReachable(source_x, source_y, dest_x, dest_y)) cout << "Yes"; else cout << "No"; return 0;} |
Python3
# Python3 program to implement# the above approach# Check if (x2, y2) can be reached# from (x1, y1)def isReachable(x1, y1, x2, y2): while(x2 > x1 and y2 > y1): # Reduce x2 by y2 until it is # less than or equal to x1 if(x2 > y2): x2 %= y2 # Reduce y2 by x2 until it is # less than or equal to y1 else: y2 %= x2 # If x2 is reduced to x1 if(x2 == x1): # Check if y2 can be # reduced to y1 or not return (y2 - y1) >= 0 and ( y2 - y1) % x1 == 0 # If y2 is reduced to y1 elif(y2 == y1): # Check if x2 can be # reduced to x1 or not return (x2 - x1) >= 0 and ( x2 - x1) % y1 == 0 else: return 0# Driver Codesource_x = 2source_y = 10dest_x = 26dest_y = 12# Function callif(isReachable(source_x, source_y, dest_x, dest_y)): print("Yes")else: print("No")# This code is contributed by Shivam Singh |
Java
// Java program to implement// the above approachclass GFG{// Check if (x2, y2) can be reached// from (x1, y1)static boolean isReachable(long x1, long y1, long x2, long y2){ while (x2 > x1 && y2 > y1) { // Reduce x2 by y2 until it is // less than or equal to x1 if (x2 > y2) x2 %= y2; // Reduce y2 by x2 until it is // less than or equal to y1 else y2 %= x2; } // If x2 is reduced to x1 if (x2 == x1) // Check if y2 can be // reduced to y1 or not return (y2 - y1) >= 0 && (y2 - y1) % x1 == 0; // If y2 is reduced to y1 else if (y2 == y1) // Check if x2 can be // reduced to x1 or not return (x2 - x1) >= 0 && (x2 - x1) % y1 == 0; else return false;}// Driver Codepublic static void main(String[] args){ long source_x = 2, source_y = 10; long dest_x = 26, dest_y = 12; if (isReachable(source_x, source_y, dest_x, dest_y)) System.out.print("Yes"); else System.out.print("No");}}// This code is contributed by shikhasingrajput |
C#
// C# program to implement// the above approachusing System;class GFG{// Check if (x2, y2) can be reached// from (x1, y1)static bool isReachable(long x1, long y1, long x2, long y2){ while (x2 > x1 && y2 > y1) { // Reduce x2 by y2 // until it is less // than or equal to x1 if (x2 > y2) x2 %= y2; // Reduce y2 by x2 // until it is less // than or equal to y1 else y2 %= x2; } // If x2 is reduced to x1 if (x2 == x1) // Check if y2 can be // reduced to y1 or not return (y2 - y1) >= 0 && (y2 - y1) % x1 == 0; // If y2 is reduced to y1 else if (y2 == y1) // Check if x2 can be // reduced to x1 or not return (x2 - x1) >= 0 && (x2 - x1) % y1 == 0; else return false;}// Driver Codepublic static void Main(String[] args){ long source_x = 2, source_y = 10; long dest_x = 26, dest_y = 12; if (isReachable(source_x, source_y, dest_x, dest_y)) Console.Write("Yes"); else Console.Write("No");}}// This code is contributed by shikhasingrajput |
Javascript
<script>// JavaScript program for//the above approach// Check if (x2, y2) can be reached// from (x1, y1)function isReachable(x1, y1, x2, y2){ while (x2 > x1 && y2 > y1) { // Reduce x2 by y2 until it is // less than or equal to x1 if (x2 > y2) x2 %= y2; // Reduce y2 by x2 until it is // less than or equal to y1 else y2 %= x2; } // If x2 is reduced to x1 if (x2 == x1) // Check if y2 can be // reduced to y1 or not return (y2 - y1) >= 0 && (y2 - y1) % x1 == 0; // If y2 is reduced to y1 else if (y2 == y1) // Check if x2 can be // reduced to x1 or not return (x2 - x1) >= 0 && (x2 - x1) % y1 == 0; else return false;} // Driver Code let source_x = 2, source_y = 10; let dest_x = 26, dest_y = 12; if (isReachable(source_x, source_y, dest_x, dest_y)) document.write("Yes"); else document.write("No"); </script> |
Output:
Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
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