First collision point of two series

Given five numbers a, b, c, d and n (where a, b, c, d, n > 0). These values represent n terms of two series. The two series formed by these four numbers are b, b+a, b+2a….b+(n-1)a and d, d+c, d+2c, ….. d+(n-1)c
These two series will collide when at any single point summation values becomes exactly the same for both the series.Print the collision point.

Example:
Input : a = 20, b = 2, 
        c = 9, d = 19, 
        n = 100
Output: 82
Explanation: 
Series1 = (2, 22, 42, 62, 82, 102...)  
Series2 = (28, 37, 46, 55, 64, 73, 82, 91..) 
So the first collision point is 82.   

A naive approach is to calculate both the series in two different arrays, and then check for each element if it collides by running two nested loops



Time complexity: O(n * n)
Auxiliary Space: O(n)

An efficient Approach to the problem mentioned above is:

* Generate all elements of first series. Let current element be x.
* If x is also an element of second series, then following conditions should satisfy.
…..a) x should be greater than or equal to first element of second series.
…..a) Difference between x and first element should be divisible by c.
*If the above conditions are satisfied then the i-th value is the required meeting point.

Below is the implementation of the above problem :

C++

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// CPP program to calculate the colliding
// point of two series
#include<bits/stdc++.h>
using namespace std;
  
  
void point(int a, int b, int c, int d, int n)
{
    int x , flag = 0;
  
    // Iterating through n terms of the 
    // first series
    for (int i = 0; i < n; i++)
    {
          
        // x is i-th term of first series
        x = b + i * a;     
          
        // d is first element of second
        // series and c is common difference
        // for second series.
        if ((x - d) % c == 0 and x - d >= 0)
        {
            cout << x << endl ;
            flag = 1;
            break;
        }
      
    }
  
    // If no term of first series is found     
    if(flag == 0)
    {
            cout << "No collision point" << endl;
    }
  
      
      
  
// Driver function
int main()
{
    int a = 20 ;
    int b = 2 ;
    int c = 9;
    int d = 19;
    int n = 20;
    point(a, b, c, d, n);
    return 0;
}
  
// This code is contributed by  'saloni1297'.

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Java

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// Java program to calculate the colliding
// point of two series
import java.io.*;
  
class GFG 
{
    static void point(int a, int b, int c, int d, int n)
    {
        int x , flag = 0;
  
        // Iterating through n terms of the 
        // first series
        for (int i = 0; i < n; i++)
        {
              
            // x is i-th term of first series
            x = b + i * a; 
              
            // d is first element of second
            // series and c is common difference
            // for second series.
            if ((x - d) % c == 0 && x - d >= 0)
            {
                System.out.println( x ) ;
                flag = 1;
                break;
            }
          
        }
      
        // If no term of first series is found 
        if(flag == 0)
        {
            System.out.println ("No collision point");
        }
      
          
    
          
      
    // Driver function
    public static void main (String[] args) 
    {
        int a = 20 ;
        int b = 2 ;
        int c = 9;
        int d = 19;
        int n = 20;
        point(a, b, c, d, n);   
      
    }
}
// This code is contributed by vt_m

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Python

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# Function to calculate the colliding point
# of two series
def point(a, b, c, d, n):
  
    # Iterating through n terms of the 
    # first series
    for i in range(n):
          
        # x is i-th term of first series
        x = b + i*a     
          
        # d is first element of second
        # series and c is common difference
        # for second series.
        if (x-d)%c == 0 and x-d >= 0:
            print
            return
  
    # If no term of first series is found      
    else:
        print "No collision point"    
     
  
# Driver code
a = 20
b = 2
c = 9
d = 19
n = 20
point(a, b, c, d, n)

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C#

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// C# program to calculate the colliding
// point of two series
using System;
  
class GFG {
      
    static void point(int a, int b, int c,
                             int d, int n)
    {
        int x, flag = 0;
  
        // Iterating through n terms of the
        // first series
        for (int i = 0; i < n; i++) {
  
            // x is i-th term of first series
            x = b + i * a;
  
            // d is first element of second
            // series and c is common difference
            // for second series.
            if ((x - d) % c == 0 && x - d >= 0) {
                Console.WriteLine(x);
                flag = 1;
                break;
            }
        }
  
        // If no term of first series is found
        if (flag == 0) {
            Console.WriteLine("No collision point");
        }
    }
  
    // Driver function
    public static void Main()
    {
        int a = 20;
        int b = 2;
        int c = 9;
        int d = 19;
        int n = 20;
        point(a, b, c, d, n);
    }
}
  
// This code is contributed by vt_m

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PHP

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<?php
// PHP program to calculate
// the colliding point of
// two series
  
  
function point($a, $b, $c, $d, $n)
{
    $x ; $flag = 0;
  
    // Iterating through 
    // n terms of the 
    // first series
    for ($i = 0; $i < $n; $i++)
    {
          
        // x is i-th term
        // of first series
        $x = $b + $i * $a
          
        // d is first element of
        // second series and c is
        // common difference for
        // second series.
        if (($x - $d) % $c == 0 and 
                      $x - $d >= 0)
        {
            echo $x;
            $flag = 1;
            break;
        }
      
    }
  
    // If no term of first 
    // series is found 
    if($flag == 0)
    {
            echo "No collision po$";
    }
  
      
      
    // Driver Code
    $a = 20 ;
    $b = 2 ;
    $c = 9;
    $d = 19;
    $n = 20;
    point($a, $b, $c, $d, $n);
  
// This code is contributed by anuj_67.
?>

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Output :

82

Time complexity : O(n)

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Improved By : vt_m



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