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 = 4s1 = 4p2 = 8s2 = 2if(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 neveropen and would like to contribute, you can also write an article using write.neveropen.co.za or mail your article to review-team@neveropen.co.za. See your article appearing on the neveropen main page and help other Geeks.
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