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Find if two people ever meet after same number of jumps

  • Difficulty Level : Easy
  • Last Updated : 29 Apr, 2021

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

 

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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 following conditions are met. 
(1) Speeds are not same 
(2) Difference between speeds divide the total distance between initial points. 
 

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

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