# Number of hours after which the second person moves ahead of the first person if they travel at a given speed

• Last Updated : 05 Apr, 2021

Given three integers A, B and K. Initially, the first person was ahead of the second person by K kms. In every hour, the first person moves ahead by A kms and the second person moves ahead by B kms. The task is to print the number of hours after which the second person crosses the first. If it is impossible to do so then print -1

Examples:

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: A = 4, B = 5, K = 1
Output:
Initially, the first person was ahead by 1 km.
After 1st hour the first and second person are at the same place.
After 2nd hour the first person moves ahead of the first person by 1 km.

Input: A = 6, B = 5, K = 1
Output: -1

A naive approach is to linearly check for every hour and print the n-th hour when the second person moves ahead of the first person.
An efficient approach is to solve the problem mathematically. The number of hours will be K / (B – A) + 1 when the second person moves ahead of the first person. Since you need to cover K kms, hence the time taken will be K / (B – A) where B – A is the speed of the second person with respect to the first person. If A ≥ B then it is not possible for the second person to cross the first person.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;` `// Function to return the number of``// hours for the second person to move ahead``int` `findHours(``int` `a, ``int` `b, ``int` `k)``{``    ``if` `(a >= b)``        ``return` `-1;` `    ``// Time taken to equalize``    ``int` `time` `= k / (b - a);` `    ``// Time taken to move ahead``    ``time` `= ``time` `+ 1;` `    ``return` `time``;``}` `// Driver code``int` `main()``{``    ``int` `a = 4, b = 5, k = 1;``    ``cout << findHours(a, b, k);``    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``import` `java.io.*;` `class` `GFG``{``    ` `// Function to return the number of``// hours for the second person to move ahead``static` `int` `findHours(``int` `a, ``int` `b, ``int` `k)``{``    ``if` `(a >= b)``        ``return` `-``1``;` `    ``// Time taken to equalize``    ``int` `time = k / (b - a);` `    ``// Time taken to move ahead``    ``time = time + ``1``;` `    ``return` `time;``}` `// Driver code``public` `static` `void` `main (String[] args)``{` `        ``int` `a = ``4``, b = ``5``, k = ``1``;``        ``System.out.println (findHours(a, b, k));``}``}` `// The code is contributed by ajit..@23`

## Python3

 `# Python3 implementation of the above approach` `# Function to return the number of``# hours for the second person to move ahead``def` `findHours(a, b, k):``    ``if` `(a >``=` `b):``        ``return` `-``1` `    ``# Time taken to equalize``    ``time ``=` `k ``/``/` `(b ``-` `a)` `    ``# Time taken to move ahead``    ``time ``=` `time ``+` `1` `    ``return` `time`  `# Driver code` `a ``=` `4``b ``=` `5``k ``=` `1``print``(findHours(a, b, k))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG``{``        ` `// Function to return the number of``// hours for the second person to move ahead``static` `int` `findHours(``int` `a, ``int` `b, ``int` `k)``{``    ``if` `(a >= b)``        ``return` `-1;` `    ``// Time taken to equalize``    ``int` `time = k / (b - a);` `    ``// Time taken to move ahead``    ``time = time + 1;` `    ``return` `time;``}` `// Driver code``static` `public` `void` `Main ()``{``    ``int` `a = 4, b = 5, k = 1;``    ``Console.Write(findHours(a, b, k));``}``}` `// The code is contributed by ajit.`

## Javascript

 ``
Output:
`2`

My Personal Notes arrow_drop_up