CODEWITHSUNDEEP
java
Abdallah Sabri
Abdallah Sabri

Reputation: 53

How to solve Tortoise racing

There is a problem called: Tortoise racing.

The question is:

Two tortoises named A and B must run a race. A starts with an average speed of 720 feet per hour. Young B knows she runs faster than A, and furthermore has not finished her cabbage.

When she starts, at last, she can see that A has a 70 feet lead but B's speed is 850 feet per hour. How long will it take B to catch A?

More generally: given two speeds v1 (A's speed, integer > 0) and v2 (B's speed, integer > 0) and a lead g (integer > 0) how long will it take B to catch A?

The result will be an array [hour, min, sec] which is the time needed in hours, minutes and seconds (round down to the nearest second) or a string in some languages.

If v1 >= v2 then return nil, nothing, null, None or {-1, -1, -1} for C++, C, Go, Nim, [] for Kotlin or "-1 -1 -1".

Examples: (form of the result depends on the language)

race(720, 850, 70) => [0, 32, 18] or "0 32 18"

race(80, 91, 37) => [3, 21, 49] or "3 21 49"

I tried to solve it like this: 

  public static int[] race(int v1, int v2, int g) {

            int v3 = v2 - v1;
            double time = (double )g / (double)v3;
            int result[] = new int[3];

            if (v2 > v1) {
                if (time > 1) {
                    while (time > 10) {
                        time /= 10;
                    }
                    result[0] = (int) time;
                    result[1] = (int) ((time - result[0]) * 60);
                    result[2] = (int) ((((time - result[0]) * 60) - result[1]) * 60);
                    System.out.print(result[0] + " " + result[1] + " " + result[2]);
                } else {
                    result[0] = 0;
                    result[1] = (int) (time * 60);
                    result[2] = (int) (((time * 60) - result[1]) * 60);
                    System.out.print(result[0] + " " + result[1] + " " + result[2]);
                }
            }
            else {
                return null;
            }

            return result;
        }

but it keeps failing on test cases, could you please help me?

Upvotes: 1

Views: 2596

Answers (2)

justgivememyicecream
justgivememyicecream

Reputation: 270

Probably you could try following code, 

public static int[] race(int v1, int v2, int g){
    if(v1 >= v2)
        return null; //B will never catch A
    int speedDifference = (v2 - v1);
    int resultInSeconds = g * 3600 / speedDifference; //*3600 to get in seconds
    int[] result = {resultInSeconds/3600, resultInSeconds%3600/60, resultInSeconds%3600%60};
    return result;
}

Upvotes: 2

Shashwat
Shashwat

Reputation: 2352

Check Below code :

public class Tortoise {
    public static int[] race(int v1, int v2, int g) {
        if (v1 >= v2)
            return null;
        int seconds = (g * 3600) / (v2 - v1);
        return new int[]{seconds / 3600, (seconds % 3600) / 60, seconds % 60};
    }

    public static void main(String[] args) {
        Tortoise.race(720, 850, 70);
    }
}

Upvotes: 9

Related Questions