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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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1271 Discussions

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  • maddykumarsingh
     + 0 comments

    function catAndMouse( catAPosition , catBPosition , mousePostion ) {

    const catAStep = Math.abs(mousePostion - catAPosition); const catBStep = Math.abs( mousePostion - catBPosition );

    let difference = catAStep - catBStep;

    let result = (difference === 0 )? 'Mouse C':(difference > 0)?'Cat B':'Cat A'; return result }

  • yashdeora98294
     + 0 comments

    Here is problem solution in Python, Java, C++, C and Javascript - https://programmingoneonone.com/hackerrank-cats-and-a-mouse-problem-solution.html

  • 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";
}

  • fkaanfirat
     + 0 comments
    def catAndMouse(x, y, z):
    
    min_catA = (z-x)**2
    min_catB = (z-y)**2 
    
    if min_catA > min_catB : 
        return "Cat B"
    elif min_catB > min_catA:
        return "Cat A"
    else : 
        return "Mouse C"

  • kristjanson_ryan
     + 0 comments

    I thought using a dictionary of exclusive boolean statements would be fun

    `def catAndMouse(x, y, z):

    catA_dist = abs(x-z)
catB_dist = abs(y-z)

d = {catA_dist < catB_dist:'Cat A',
     catA_dist > catB_dist:'Cat B',
     catA_dist == catB_dist:'Mouse C',
}

return d[True]