Find if two people ever meet after same number of jumps

Two people start races from two different points p1 and p2. They cover s1 and s2 meters in a jump. Find if they will ever meet at a point after the same number of jumps.

Examples:

Input : p1 = 6, s1 = 3, 
        p2 = 8, s2 = 2
Output : Yes
Explanation: 6->9->12
             8->10->12
They meet after two jumps.

Input : p1 = 4, s1 = 4, 
        p2 = 8, s2 = 2
Output : Yes
Explanation: 4->8->12
             8->10->12

Input : p1 = 0, s1 = 2, 
        p2 = 5, s2 = 3
Output : No

Input : p1 = 42, s1 = 3, 
        p2 = 94, s2 = 2
Output : Yes

A simple solution is to make them jump one by one. After every jump, see if they are same point or not.

An efficient solution is based on below facts: 
Since starting points are always different, they will meet if the following conditions are met. 
(1) Speeds are not the same 
(2) Difference between speeds divide by the total distance between initial points. 

Implementation:

C++

// C++ program to find any one of them
// can overtake the other
#include<bits/stdc++.h>
using namespace std;

// function to find if any one of them can
// overtake the other
bool sackRace(int p1, int s1, int p2, int s2){

    // Since starting points are always
    // different, they will meet if following
    // conditions are met.
    // (1) Speeds are not same
    // (2) Difference between speeds divide the
    //     total distance between initial points.
    return ( (s1 > s2 && (p2 - p1) % (s1 - s2) == 0) ||
             (s2 > s1 && (p1 - p2) % (s2 - s1) == 0));
}

// driver program
int main()
{
    int p1 = 4, s1 = 4, p2 = 8, s2 = 2;
    sackRace(p1, s1, p2, s2)? cout << "Yes\n" :
                              cout << "No\n";
    return 0;
}

Java

// java program to find any one of them
// can overtake the other
import java.util.Arrays;

public class GFG {

    // function to find if any one of
    // them can overtake the other
    static boolean sackRace(int p1, int s1,
                            int p2, int s2)
    {

        // Since starting points are
        // always different, they will
        // meet if following conditions
        // are met.
        // (1) Speeds are not same
        // (2) Difference between speeds
        // divide the total distance
        // between initial points.
        return ( (s1 > s2 && (p2 - p1) %
                    (s1 - s2) == 0) ||
                    (s2 > s1 && (p1 - p2)
                    % (s2 - s1) == 0));
    }

    public static void main(String args[])
    {
        int p1 = 4, s1 = 4, p2 = 8, s2 = 2;

        if(sackRace(p1, s1, p2, s2))
            System.out.println("Yes" );
        else
            System.out.println("No");

    }
}

// This code is contributed by Sam007.

Python3

# python program to find any one of them
# can overtake the other

# function to find if any one of them can
# overtake the other
def sackRace(p1, s1, p2, s2):

    # Since starting points are always
    # different, they will meet if following
    # conditions are met.
    # (1) Speeds are not same
    # (2) Difference between speeds divide the
    #     total distance between initial points.
    return ( (s1 > s2 and (p2 - p1) % (s1 - s2) == 0)
         or (s2 > s1 and (p1 - p2) % (s2 - s1) == 0))


# driver program
p1 = 4
s1 = 4
p2 = 8
s2 = 2
if(sackRace(p1, s1, p2, s2)):
    print("Yes")
else:
    print("No")

# This code is contributed by Sam007

C#

// C# program to find any one of them
// can overtake the other
using System;

class GFG {

    // function to find if any one of
    // them can overtake the other
    static bool sackRace(int p1, int s1,
                          int p2, int s2)
    {

        // Since starting points are
        // always different, they will
        // meet if following conditions
        // are met.
        // (1) Speeds are not same
        // (2) Difference between speeds
        // divide the total distance
        // between initial points.
        return ( (s1 > s2 && (p2 - p1) %
                       (s1 - s2) == 0) ||
                    (s2 > s1 && (p1 - p2)
                       % (s2 - s1) == 0));
    }

    // Driver code
    public static void Main()
    {
        int p1 = 4, s1 = 4, p2 = 8,
        s2 = 2;

        if(sackRace(p1, s1, p2, s2))
            Console.WriteLine("Yes" );
        else
            Console.WriteLine("No");

    }
}

// This code is contributed by Sam007.

PHP

<?php
// PHP program to find any one of them
// can overtake the other

// function to find if any one of
// them can overtake the other
function sackRace($p1, $s1, $p2, $s2)
{

    // Since starting points are always
    // different, they will meet if following
    // conditions are met.
    // (1) Speeds are not same
    // (2) Difference between speeds divide the
    //     total distance between initial points.
    return (($s1 > $s2 && ($p2 - $p1) % ($s1 - $s2) == 0) ||
            ($s2 > $s1 && ($p1 - $p2) % ($s2 - $s1) == 0));
}

    // Driver Code
    $p1 = 4;
    $s1 = 4;
    $p2 = 8;
    $s2 = 2;
    if(sackRace($p1, $s1, $p2, $s2))
        echo "Yes\n" ;
    else
        echo "No\n";

// This code is contributed by Sam007
?>

Javascript

<script>

// JavaScript program to find any one of them
// can overtake the other

    // function to find if any one of
    // them can overtake the other
    function sackRace(p1, s1, p2, s2)
    {

        // Since starting points are
        // always different, they will
        // meet if following conditions
        // are met.
        // (1) Speeds are not same
        // (2) Difference between speeds
        // divide the total distance
        // between initial points.
        return ( (s1 > s2 && (p2 - p1) %
                    (s1 - s2) == 0) ||
                    (s2 > s1 && (p1 - p2)
                    % (s2 - s1) == 0));
    }

// Driver Code
        let p1 = 4, s1 = 4, p2 = 8, s2 = 2;

        if(sackRace(p1, s1, p2, s2))
            document.write("Yes" );
        else
            document.write("No");

// This code is contributed by susmitakundugoaldanga.
</script>

Output

Yes

Time Complexity:  O(1)
Auxiliary Space: O(1)

This article is contributed by Vishal Kumar Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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