Check if two coordinates can be made equal by incrementing/decrementing by K1 and K2 respectively
Given two integer coordinates (X1, Y1) and (X2, Y2) and two positive integers K1 and K2, the task is to check if both the coordinates can be made equal by performing the following steps any number of times:
- Add or subtract K1 from either or both coordinates of (X1, Y1).
- Add or subtract K2 from either or both coordinates of (X2, Y2).
If it is possible to make (X1, Y1) and (X2, Y2) equal, then print Yes. Otherwise, print No.
Examples:
Input: X1 = 10, Y1 = 10, X2 = 18, Y2 = 13, K1 = 3, K2 = 4
Output: Yes
Explanation:
Following are the moves that can be taken to make both the coordinates equal:
- Move point (X1, Y1) as (10, 10) -> (10, 13).
- Move point (X2, Y2) as (18, 13) -> (14, 13) -> (10, 13)
From the above operations, both the coordinates can be made equal, then print Yes.
Input: X1 = 10, Y1 = 10, X2 = 18, Y2 = 13, K1 = 10, K2 = 10
Output: No
Approach: This problem can be solved using Greedy Approach based on the observation that the moves can be taken in x-direction for (X1, Y1) point is n1 and the moves taken in x-direction for (X2, Y2) point is n2, then the expression can be written as:
n1*K1 + n2*K2 = abs(X1 – X2),
…where n1 and n2 are non negative integers.
Similarly the same can be written for y-direction as:
n3*K1 + n4*K2 = abs(Y1 – Y2),
…where n3 and n4 are non negative integers.
Now, it can be seen that, the problem has been reduced to find if the above equation has solutions or not. If both equations have non-negative solutions then print ‘Yes’. Otherwise, print ‘No’.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to check if both can be merged int twoPointsReachable( int X1, int Y1, int X2, int Y2, int K1, int K2) { // Calculate gcd of K1, K2 int g = __gcd(K1, K2); // Solve for the X-axis bool reachableOnX = 0; // Calculate distance between the // X-coordinates int X_distance = abs (X1 - X2); // Check the divisibility if (X_distance % g == 0) { reachableOnX = 1; } // Solve for the Y-axis bool reachableOnY = 0; // Calculate distance on between // X coordinates int Y_distance = abs (Y1 - Y2); // Check for the divisibility if (Y_distance % g == 0) { reachableOnY = 1; } // Check if both solutions exist if (reachableOnY && reachableOnX) { cout << "Yes" << "\n" ; } else { cout << "No" << "\n" ; } return 0; } // Driver Code int main() { // Given Input int X1 = 10, Y1 = 10, X2 = 18; int Y2 = 13, K1 = 3, K2 = 4; // Function Call twoPointsReachable(X1, X2, Y1, Y2, K1, K2); return 0; } |
Java
// Java program for the above approach import java.util.*; public class GFG{ static int __gcd( int a, int b) { return b == 0 ? a : __gcd(b, a % b); } // Function to check if both can be merged static int twoPointsReachable( int X1, int Y1, int X2, int Y2, int K1, int K2) { // Calculate gcd of K1, K2 int g = __gcd(K1, K2); // Solve for the X-axis boolean reachableOnX = (g == 0 ); // Calculate distance between the // X-coordinates int X_distance = Math.abs(X1 - X2); // Check the divisibility if (X_distance % g == 0 ) { reachableOnX = (g == 1 ); } // Solve for the Y-axis boolean reachableOnY = (g == 0 ); // Calculate distance on between // X coordinates int Y_distance = Math.abs(Y1 - Y2); // Check for the divisibility if (Y_distance % g == 0 ) { reachableOnY = (g == 1 ); } // Check if both solutions exist if (reachableOnY && reachableOnX) { System.out.print( "Yes" ); } else { System.out.print( "No" ); } return 0 ; } // Driver Code public static void main (String[] args) { // Given Input int X1 = 10 , Y1 = 10 , X2 = 18 ; int Y2 = 13 , K1 = 3 , K2 = 4 ; // Function Call twoPointsReachable(X1, X2, Y1, Y2, K1, K2); } } // This code is contributed by shivanisinghss2110 |
Python3
# python program for the above approach # Function to check if both can be merged from math import * def twoPointsReachable(X1, Y1, X2, Y2, K1, K2): # Calculate gcd of K1, K2 g = gcd(K1, K2) # Solve for the X-axis reachableOnX = 0 # Calculate distance between the # X-coordinates X_distance = abs (X1 - X2) # Check the divisibility if (X_distance % g = = 0 ): reachableOnX = 1 # Solve for the Y-axis reachableOnY = 0 # Calculate distance on between # X coordinates Y_distance = abs (Y1 - Y2) # Check for the divisibility if (Y_distance % g = = 0 ): reachableOnY = 1 # Check if both solutions exist if (reachableOnY and reachableOnX): print ( "Yes" ) else : print ( "No" ) return 0 # Driver Code # Given Input X1 = 10 Y1 = 10 X2 = 18 Y2 = 13 K1 = 3 K2 = 4 # Function Call twoPointsReachable(X1, X2, Y1, Y2, K1, K2) # This code is contributed by anudeep23042002 |
C#
// C++ program for the above approach using System; public class GFG { static int __gcd( int a, int b) { return b == 0 ? a : __gcd(b, a % b); } // Function to check if both can be merged static int twoPointsReachable( int X1, int Y1, int X2, int Y2, int K1, int K2) { // Calculate gcd of K1, K2 int g = __gcd(K1, K2); // Solve for the X-axis bool reachableOnX = Convert.ToBoolean(0); // Calculate distance between the // X-coordinates int X_distance = Math.Abs(X1 - X2); // Check the divisibility if (X_distance % g == 0) { reachableOnX = Convert.ToBoolean(1); } // Solve for the Y-axis bool reachableOnY = Convert.ToBoolean(0); // Calculate distance on between // X coordinates int Y_distance = Math.Abs(Y1 - Y2); // Check for the divisibility if (Y_distance % g == 0) { reachableOnY = Convert.ToBoolean(1); } // Check if both solutions exist if (reachableOnY && reachableOnX) { Console.Write( "Yes" ); } else { Console.Write( "No" ); } return 0; } // Driver Code public static void Main(String[] args) { // Given Input int X1 = 10, Y1 = 10, X2 = 18; int Y2 = 13, K1 = 3, K2 = 4; // Function Call twoPointsReachable(X1, X2, Y1, Y2, K1, K2); } } // This code is contributed by shivanisinghss2110 |
Javascript
<script> // JavaScript Program to implement // the above approach function __gcd(a, b) { // Everything divides 0 if (a == 0) return b; if (b == 0) return a; // base case if (a == b) return a; // a is greater if (a > b) return __gcd(a - b, b); return __gcd(a, b - a); } // Function to check if both can be merged function twoPointsReachable(X1, Y1, X2, Y2, K1, K2) { // Calculate gcd of K1, K2 let g = __gcd(K1, K2); // Solve for the X-axis let reachableOnX = 0; // Calculate distance between the // X-coordinates let X_distance = Math.abs(X1 - X2); // Check the divisibility if (X_distance % g == 0) { reachableOnX = 1; } // Solve for the Y-axis let reachableOnY = 0; // Calculate distance on between // X coordinates let Y_distance = Math.abs(Y1 - Y2); // Check for the divisibility if (Y_distance % g == 0) { reachableOnY = 1; } // Check if both solutions exist if (reachableOnY && reachableOnX) { document.write( "Yes" + "<br>" ); } else { document.write( "No" + "<br>" ); } return 0; } // Driver Code // Given Input let X1 = 10, Y1 = 10, X2 = 18; let Y2 = 13, K1 = 3, K2 = 4; // Function Call twoPointsReachable(X1, X2, Y1, Y2, K1, K2); // This code is contributed by Potta Lokesh </script> |
Yes
Time Complexity: O(log(max(K1, K2)))
Auxiliary Space: O(1)
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