# 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``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

 ` ``\$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

 ``

Output :

`Yes`

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