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 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++
#include<bits/stdc++.h>
using namespace std;
bool sackRace( int p1, int s1, int p2, int s2){
return ( (s1 > s2 && (p2 - p1) % (s1 - s2) == 0) ||
(s2 > s1 && (p1 - p2) % (s2 - s1) == 0));
}
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
import java.util.Arrays;
public class GFG {
static boolean sackRace( int p1, int s1,
int p2, int s2)
{
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" );
}
}
|
Python3
def sackRace(p1, s1, p2, s2):
return ( (s1 > s2 and (p2 - p1) % (s1 - s2) = = 0 )
or (s2 > s1 and (p1 - p2) % (s2 - s1) = = 0 ))
p1 = 4
s1 = 4
p2 = 8
s2 = 2
if (sackRace(p1, s1, p2, s2)):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG {
static bool sackRace( int p1, int s1,
int p2, int s2)
{
return ( (s1 > s2 && (p2 - p1) %
(s1 - s2) == 0) ||
(s2 > s1 && (p1 - p2)
% (s2 - s1) == 0));
}
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" );
}
}
|
PHP
<?php
function sackRace( $p1 , $s1 , $p2 , $s2 )
{
return (( $s1 > $s2 && ( $p2 - $p1 ) % ( $s1 - $s2 ) == 0) ||
( $s2 > $s1 && ( $p1 - $p2 ) % ( $s2 - $s1 ) == 0));
}
$p1 = 4;
$s1 = 4;
$p2 = 8;
$s2 = 2;
if (sackRace( $p1 , $s1 , $p2 , $s2 ))
echo "Yes\n" ;
else
echo "No\n" ;
?>
|
Javascript
<script>
function sackRace(p1, s1, p2, s2)
{
return ( (s1 > s2 && (p2 - p1) %
(s1 - s2) == 0) ||
(s2 > s1 && (p1 - p2)
% (s2 - s1) == 0));
}
let p1 = 4, s1 = 4, p2 = 8, s2 = 2;
if (sackRace(p1, s1, p2, s2))
document.write( "Yes" );
else
document.write( "No" );
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Last Updated :
16 Feb, 2023
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