Hackerrank | Cats and a Mouse

Two cats and a mouse are at various positions on a line. You will be given their starting positions. Your task is to determine which cat will reach the mouse first, assuming the mouse does not move and the cats travel at equal speed. If the cats arrive at the same time, the mouse will be allowed to move and it will escape while they fight. You are given q queries in the form of x,y , and z representing the respective positions for cats A and B, and for mouse C. Complete the function to return the appropriate answer to each query, which will be printed on a new line. If cat catches the mouse first, print Cat A. If cat catches the mouse first, print Cat B. If both cats reach the mouse at the same time, print Mouse C as the two cats fight and mouse escapes.

You are given q queries in the form of x, y, zand  representing the respective positions for cats A andB , and for mouse C . Complete the function  to return the appropriate answer to each query, which will be printed on a new line.

• If cat A catches the mouse first, print Cat A.
• If cat B catches the mouse first, print Cat B.
• If both cats reach the mouse at the same time, print Mouse C as the two cats fight and mouse escapes.

Example

x=2

y=5

z=4

The cats are at positions2  (Cat A) and 5  (Cat B), and the mouse is at position 4. Cat B, at position 5 will arrive first since it is only 1 unit away while the other is 2 units away. Return ‘Cat B’.

Function Description

Complete the catAndMouse function in the editor below.

catAndMouse has the following parameter(s):

• int x: Cat ‘s position
• int y: Cat ‘s position
• int z: Mouse ‘s position

Returns

• string: Either ‘Cat A’, ‘Cat B’, or ‘Mouse C’

Input Format

The first line contains a single integer, , denoting the number of queries.
Each of the  subsequent lines contains three space-separated integers describing the respective values of  (cat ‘s location),  (cat ‘s location), and  (mouse ‘s location).

Constraints

• 1<=q<=100
• 1<=x,y,z<=100

Sample Input 0

2
1 2 3
1 3 2

Sample Output 0

Cat B
Mouse C

Explanation 0

Query 0: The positions of the cats and mouse are shown below: 

Cat  will catch the mouse first, so we print Cat B on a new line.

Query 1: In this query, cats  and  reach mouse  at the exact same time: 

Because the mouse escapes, we print Mouse C on a new line.

Python Implementation

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the catAndMouse function below.
def catAndMouse(x, y, z):
    distance_cat_a = abs(x - z)
    distance_cat_b = abs(y - z)

    if distance_cat_a < distance_cat_b:
        return "Cat A"
    elif distance_cat_a > distance_cat_b:
        return "Cat B"
    else:
        return "Mouse C"

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    q = int(input())

    for q_itr in range(q):
        xyz = input().split()

        x = int(xyz[0])

        y = int(xyz[1])

        z = int(xyz[2])

        result = catAndMouse(x, y, z)

        fptr.write(result + '\n')

    fptr.close()

C Implementation

#include <stdio.h>
#include <stdlib.h>

char* catAndMouse(int x, int y, int z) {
    int distance_cat_a = abs(x - z);
    int distance_cat_b = abs(y - z);

    if (distance_cat_a < distance_cat_b) {
        return "Cat A";
    } else if (distance_cat_a > distance_cat_b) {
        return "Cat B";
    } else {
        return "Mouse C";
    }
}

int main() {
    // Example usage:
    int queries[][3] = {{1, 2, 3}, {1, 3, 2}, {2, 1, 3}};
    int num_queries = sizeof(queries) / sizeof(queries[0]);

    for (int i = 0; i < num_queries; i++) {
        char* result = catAndMouse(queries[i][0], queries[i][1], queries[i][2]);
        printf("%s\n", result);
    }

    return 0;
}