Cats and a Mouse


Problem Statement :


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 catAndMouse 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 positions 2 (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 A's position
int y: Cat B's position
int z: Mouse C's position

Returns

string: Either 'Cat A', 'Cat B', or 'Mouse C'


Input Format

The first line contains a single integer, q, denoting the number of queries.
Each of the q subsequent lines contains three space-separated integers describing the respective values of x (cat A's location), y (cat B's location), and z (mouse C's location).


Constraints
1 <= q <= 100
1 <= x,y,z <= 100


Solution :



title-img 


                            Solution in C :

python 3  :

#!/bin/python3

import sys


q = int(input().strip())
for a0 in range(q):
    x,y,z = input().strip().split(' ')
    x,y,z = [int(x),int(y),int(z)]
    
    if abs(x - z) < abs(y - z):
        print('Cat A')
    elif abs(x - z) > abs(y - z):
        print('Cat B')
    else:
        print('Mouse C')
    






java  :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int q = in.nextInt();
        for(int a0 = 0; a0 < q; a0++){
            int x = in.nextInt();
            int y = in.nextInt();
            int z = in.nextInt();
            
            int dist = Math.abs(x-z);
            int dist2 = Math.abs(y-z);
            if (dist < dist2) System.out.println("Cat A");
            else if (dist > dist2) System.out.println("Cat B");
            else System.out.println("Mouse C");
        }
    }
}











C++  :

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int Q;
    scanf("%d", &Q);
    while(Q--)
    {
        int x, y, z;
        scanf("%d%d%d", &x, &y, &z);
        x=abs(x-z);
        y=abs(y-z);
        if(x==y)
            printf("Mouse C\n");
        else if(x<y)
            printf("Cat A\n");
        else
            printf("Cat B\n");
    }
    return 0;
}








C  :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
    int q; 
    scanf("%d",&q);
    for(int a0 = 0; a0 < q; a0++){
        int x; 
        int y; 
        int z; 
        scanf("%d %d %d",&x,&y,&z);
        int a = abs(x - z);
        int b = abs(y - z);
        if (a < b) printf("Cat A\n");
        else if (b < a) printf("Cat B\n");
        else printf("Mouse C\n");
    }
    return 0;
}








View More Similar Problems

Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →

Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

View Solution →

Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

View Solution →

Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

View Solution →

Cycle Detection

A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer

View Solution →

Find Merge Point of Two Lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share

View Solution →