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++ 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 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 |
# 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++ 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 |
<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)