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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Cats and a Mouse
  5. Discussions

Cats and a Mouse

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  • rahulvictorsunk1
     + 0 comments

    A Short and Easy C# Solution using Ternary operator With Time and Space Complexity of O(1).

    static string catAndMouse(int x, int y, int z) 
{
	int a = Math.Abs(x - z), b = Math.Abs(y - z);
	return (a < b) ? "Cat A" : (a == b) ? "Mouse C" : "Cat B";
}

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/vbV5-DqJU74

    string catAndMouse(int x, int y, int z) {
    int a = abs(x-z), b = abs(y-z);
    if(a==b) return "Mouse C";
    if(a < b) return "Cat A";
    return "Cat B";
}

    Short version

    string catAndMouse(int x, int y, int z) {
    int a = abs(x-z), b = abs(y-z);
    return (a > b) ? "Cat B" : (a < b) ? "Cat A" : "Mouse C";
}

  • kaadir_yaalcin
     + 0 comments

    Here is the solution for C#

    int res = Comparer.Default.Compare(Math.Abs(x-z),Math.Abs(y-z));
    return res < 0 ? "Cat A" : res > 0 ? "Cat B" : "Mouse C";

  • kunalkawate424
     + 0 comments

    Solution from my side for this problem

    def catAndMouse(x, y, z):
    a = abs(x-z)
    b = abs(y-z)
    if(a == b):
        return 'Mouse C'
    elif(a<b):
        return 'Cat A'
    elif(a>b):
        return 'Cat B'

  • dhruvbhagat98
     + 0 comments

    JS, Javascript Solution:-

    function catAndMouse(x, y, z) {
    const aMDistance = Math.abs(x-z)
    const bMDistance = Math.abs(y-z)
    if (aMDistance === bMDistance) return "Mouse C"; else if (aMDistance > bMDistance) return 'Cat B';
    return 'Cat A';
}