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  1. Gaming Array 1
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Gaming Array 1

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32 Discussions

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  • matteo_monaco201
     + 0 comments

    Python 3

    def gamingArray(arr):
    last = 1
    curr_max = arr[0]
    for number in arr[1:]:
        if number > curr_max:
            curr_max = number
            last += 1
    return "ANDY" if last % 2 == 0 else "BOB"

  • alihopyar
     + 0 comments

    Here is my solution to handle large inputs in Java;

    private static TreeMap<Integer, Integer> indexMaxMap = new TreeMap<>();
    
    private static void createIndexMaxMap(List<Integer> arr) {
        indexMaxMap.clear();
        for (int i = 0; i < arr.size(); i++) {
            indexMaxMap.put(arr.get(i), i);
        }
    }
    
    private static List<Integer> getNewList(List<Integer> arr) {
        int lastIndex = arr.size() - 1;
        while(indexMaxMap.lastEntry().getValue() > lastIndex) {
            indexMaxMap.remove(indexMaxMap.lastKey());
        }
        
        int index = indexMaxMap.remove(indexMaxMap.lastKey());
        System.out.println("Last index: " + index);
        return arr.subList(0, index);
        
    }
    
    /*
     * Complete the 'gamingArray' function below.
     *
     * The function is expected to return a STRING.
     * The function accepts INTEGER_ARRAY arr as parameter.
     */

    public static String gamingArray(List<Integer> arr) {
        createIndexMaxMap(arr);
        int moveCount = 0;
        
        while (arr.size() > 0) {
            arr = getNewList(arr);
            moveCount++;
        }
        
        if (moveCount % 2 == 1) {
            return "BOB";
        }
        
        return "ANDY";
    }

  • vadym_usatiuk
     + 0 comments

    Java, all test cases

     public static String gamingArray(List<Integer> arr) {
    int count = 0;
        int bigger = 0;
        
        for(Integer i : arr){
            if(i > bigger){
                bigger = i;
                count++;
            }    
        }

        if (count % 2 == 0) {
            return "ANDY";
        }

        return "BOB";
}
}

  • bernacikkamil
     + 1 comment

    python3 slow version:

    def gamingArray(arr):
    number_of_moves = 0
    while len(arr) != 0:
        number_of_moves += 1
        arr = arr[:arr.index(max(arr))]
    if number_of_moves % 2 == 0:
        return "ANDY"
    return "BOB"

    fast version:

    def gamingArray(arr):
    number_of_moves = 0
    moves = [(0, arr[0])]
    for i, number in enumerate(arr[1:]):
        if number > moves[-1][1]:
            moves.append((i+1, number))
    if len(moves) % 2 == 0:
        return "ANDY"
    return "BOB"

  • michaelnabatov1
     + 0 comments

    JS:

    function gamingArray(arr) {
    const getMaxIdx = (maxIdx)=>{
        let max = 0;
        for (let i = 0; i < maxIdx; i++){
            if (arr[i] > max) max = arr[i];
        }
        return arr.indexOf(max);
    }
    
    let cnt = 0;
    let idx = arr.length;
    while (idx > 0){
        idx = getMaxIdx(idx); 
        cnt++;
    }
    return (cnt%2 === 0) ? 'ANDY' : 'BOB'
}