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# Find if two people ever meet after same number of jumps

• Difficulty Level : Easy
• Last Updated : 16 Feb, 2023

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``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
```

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.