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Check if two coordinates can be made equal by incrementing/decrementing by K1 and K2 respectively

  • Difficulty Level : Hard
  • Last Updated : 24 Aug, 2021
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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:

  1. Move point (X1, Y1) as (10, 10) -> (10, 13).
  2. 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>

 
 

Output: 
Yes

 

 

Time Complexity: O(log(max(K1, K2)))
Auxiliary Space: O(1)

 

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