Check if a destination is reachable from source with two movements allowed | Set 2 - GeeksforGeeks

07 May, 2021
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)
(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: Yes
Explanation: 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++
Python3
Java
C#
Javascript
// 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 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 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# 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
<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>
Yes
Time Complexity: O(1) Auxiliary Space: O(1) 
SHIVAMSINGH67
shikhasingrajput
souravghosh0416
interview-preparation
MakeMyTrip
Modular Arithmetic
Greedy
Mathematical
MakeMyTrip
Greedy
Mathematical
Modular Arithmetic
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Huffman Coding | Greedy Algo-3
Coin Change | DP-7
Fractional Knapsack Problem
Activity Selection Problem | Greedy Algo-1
Program for Shortest Job First (or SJF) CPU Scheduling | Set 1 (Non- preemptive)
Program for Fibonacci numbers
C++ Data Types
Set in C++ Standard Template Library (STL)
Coin Change | DP-7
Merge two sorted arrays