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 :
Time complexity : O(n)
This article is contributed by Twinkle Bajaj. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to firstname.lastname@example.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.