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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Halloween Sale
  5. Discussions

Halloween Sale

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674 Discussions

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  • sivaram_movva1
     + 0 comments

    public static int howManyGames(int p, int d, int m, int s) { // Return the number of games you can buy

        int n=0;        

    while(s-p>=0)
    {            
        n++;
        s-=p;            
        p-=d;

        if(p<=m)
        {
            p=m;
        }      
    }

    return n;
}

  • greivin_lopez
     + 0 comments
    def how_many_games(p, d, m, s):
    if s < p:
        return 0
    c = 0
    while s >= m:
        s -= (p - c*d) if ((p - c*d) >= m) else m
        c += 1 if s >= 0 else 0
    return c

  • yashdeora98294
     + 0 comments

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

  • unknownagnibha
     + 0 comments

    PYTHON If in dought pls ask

    def howManyGames(p, d, m, s):
    # Return the number of games you can buy
    games=0
    i=0
    while s>=0:
        cost=p-d*i
        if cost>m:
            s-=cost
        else:
            s-=m
        games+=1
        i+=1
    return games-1

  • alban_tyrex
     + 1 comment

    Here are my approaches in c++, you can see the explanation here : https://youtu.be/VFVaLMRzhV4

    Approach 1 O(n)

    int howManyGames(int p, int d, int m, int s) {
    int result = 0;
    while( s >= p){
        s -= p;
        p = max(m, p-d);
        result++;
    }
    return result;
}

    Approach 2 O(n) with recursion

    int howManyGames(int p, int d, int m, int s){
    if( s < p) return 0;
    return howManyGames( max(m, p-d), d, m, s - p) + 1;
}

    Approach 3 O(1), incomplete solution (failure of a test case), feel free to comment to share a way to complete it.

    int howManyGames(int p, int d, int m, int s) {
    int n = 1 + (p - m) / d;
    int total = (p + p - (n - 1) * d) * n / 2;
    if(total <= s) {
        return n + (s - total) / m;
    }
    else 
        return 0;
        // this return need to be computed properly
}