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Number of Distinct Meeting Points on a Circular Road
  • Difficulty Level : Easy
  • Last Updated : 31 Jul, 2018

Consider two cars A and B, running infinitely (either clockwise or anti-clockwise) on a circular road. Given the speed of both the cars a and b. If a or b is positive, indicate they are moving in clockwise, Else they are moving in the anti-clockwise direction. The task is to find the number of distinct points they will meet each other at.

Examples:

Input : a = 1, b = -1
Output : 2

Explanation
Car A is moving clockwise while Car B is moving anti-clockwise but their
speeds are same, so they will meet at two points i.e at the starting point
and diametrically corresponding opposite point on the road.

Input : a = 1, b = 2
Output : 1

Pre-Requisites: GCD | LCM

Approach :
Let the circumference of the circular road be d.
Let the time taken by cars A and B be ta and tb respectively. Their relative speed is a – b.

A and B start from the starting point and after some time, they will meet at starting point again. This time can be calculated by the LCM of ta and tb. Within this time period, they may meet at certain points which needs to be found out. Observe that, after they meet at the starting point they keep on meeting at the same point.



Time taken to meet again at the starting point will be,

T1 = LCM(ta, tb) = LCM(d/a, d/b) = d/GCD(a, b)

Let them meet N times in the time period T1.
So, the time delay between their consecutive meets is, say T2 can be calculated as,

T2 = (T1 / N).

This time can be calculated by calculating the time taken to meet for the first time after they start.
So, the time taken by them to meet for the first time,
Therefore, T2 = (d / (a – b)).

Dividing T1 by T2, we get,
N = (T1 / T2) = ((a – b) / GCD(a, b))

Below is implementation for the above approach:

C++




// CPP Program to find number of distinct point of meet on a circular road
#include <bits/stdc++.h>
using namespace std;
  
// Returns the GCD of two number.
int gcd(int a, int b)
{
    int c = a % b;
    while (c != 0) {
        a = b;
        b = c;
        c = a % b;
    }
    return b;
}
  
// Returns the number of distinct meeting points.
int numberOfmeet(int a, int b)
{
    int ans;
  
    // Find the relative speed.
    if (a > b)
        ans = a - b;
    else
        ans = b - a;
  
    // convert the negative value to positive.
    if (a < 0)
        a = a * (-1);
  
    if (b < 0)
        b = b * (-1);
  
    return ans / gcd(a, b);
}
  
// Driver Code
int main()
{
    int a = 1, b = -1;
  
    cout << numberOfmeet(a, b) << endl;
    return 0;
}

Java




// Java Program to find number 
// of distinct point of meet 
// on a circular road
import java.io.*;
  
class GFG 
{
      
// Returns the GCD 
// of two number.
static int gcd(int a, int b)
{
    int c = a % b;
    while (c != 0
    {
        a = b;
        b = c;
        c = a % b;
    }
    return b;
}
  
// Returns the number of
// distinct meeting points.
static int numberOfmeet(int a, 
                        int b)
{
    int ans;
  
    // Find the relative speed.
    if (a > b)
        ans = a - b;
    else
        ans = b - a;
  
    // convert the negative 
    // value to positive.
    if (a < 0)
        a = a * (-1);
  
    if (b < 0)
        b = b * (-1);
  
    return ans / gcd(a, b);
}
  
// Driver Code
public static void main (String[] args)
{
    int a = 1, b = -1;
    System.out.println(numberOfmeet(a, b));
}
}
  
// This code is contributed by @ajit

Python3




# Python3 Program to find 
# number of distinct point 
# of meet on a circular road
import math
  
# Returns the number of
# distinct meeting points.
def numberOfmeet(a, b):
    ans = 0;
      
    # Find the relative speed.
    if (a > b):
        ans = a - b;
    else:
        ans = b - a;
          
    # convert the negative
    # value to positive.
    if (a < 0):
        a = a * (-1);
    if (b < 0):
        b = b * (-1);
    return int(ans / math.gcd(a, b));
  
# Driver Code
a = 1;
b = -1;
print(numberOfmeet(a, b));
  
# This code is contributed by mits

C#




// C# Program to find number 
// of distinct point of meet 
// on a circular road
using System;
  
class GFG
{
  
// Returns the GCD 
// of two number.
static int gcd(int a, int b)
{
    int c = a % b;
    while (c != 0) 
    {
        a = b;
        b = c;
        c = a % b;
    }
    return b;
}
  
// Returns the number of
// distinct meeting points.
static int numberOfmeet(int a, 
                        int b)
{
    int ans;
  
    // Find the relative speed.
    if (a > b)
        ans = a - b;
    else
        ans = b - a;
  
    // convert the negative 
    // value to positive.
    if (a < 0)
        a = a * (-1);
  
    if (b < 0)
        b = b * (-1);
  
    return ans / gcd(a, b);
}
  
// Driver Code
static public void Main ()
{
    int a = 1, b = -1;
    Console.WriteLine(
            numberOfmeet(a, b));
}
}
  
// This code is contributed 
// by @ajit

PHP




<?php
// PHP Program to find number
// of distinct point of meet
// on a circular road
  
// Returns the GCD of two number.
function gcd($a, $b)
{
    $c = $a % $b;
    while ($c != 0)
    {
        $a = $b;
        $b = $c;
        $c = $a % $b;
    }
    return $b;
}
  
// Returns the number of
// distinct meeting points.
function numberOfmeet($a, $b)
{
    $ans;
  
    // Find the relative speed.
    if ($a > $b)
        $ans = $a - $b;
    else
        $ans = $b - $a;
  
    // convert the negative
    // value to positive.
    if ($a < 0)
        $a = $a * (-1);
  
    if ($b < 0)
        $b = $b * (-1);
  
    return $ans / gcd($a, $b);
}
  
// Driver Code
$a = 1;
$b = -1;
  
echo numberOfmeet($a, $b)."\n";
      
// This code is contributed by mits
?>


Output:
2

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