# 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 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; ` `} ` |

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## 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. ` |

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## 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 ` |

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## 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. ` |

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## 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 ` `?> ` |

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**Output :**

Yes

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

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