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

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:

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`

Time Complexity: O(1), since there is a basic arithmetic operation that takes constant time.
Auxiliary Space: O(1), since no extra space has been taken.

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