# 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: 2
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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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. `

Output:

```2
```

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Improved By : mohit kumar 29, jit_t

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