# 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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