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

Halloween Sale

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672 Discussions

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  • 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
}

  • akash_prajapati6
     + 0 comments
    def howManyGames(p, d, m, s):
    total=0
    count=0
    while True:
        if(total+p>s):
            return count
        else:
            total+=p
            count+=1
        if(p-d<m):
            p=m
        else:
            p-=d

  • praveen_de_silva
     + 0 comments

    Time Complexity : O(1)

    Visit my code in github : https://github.com/praveen-de-silva/HackerRank_ProblemSolving/blob/main/HaloweenSale.cpp

    int howManyGames(int p, int d, int m, int s) { // invalid condition if (s < p) { return 0; }

    int n1 = 1 + (p - m) / d; // max number of games down to 'm'
int s1 = n1 * (2*p - (n1 - 1) * d) / 2;      // total cost down to 'm'
int res;

// case 01 : 's' is in between 'p' and 's1'
if (s < s1) {
    double det = sqrt((2*p + d) * (2*p + d) - 8*s*d);
    res = (2*p + d - det) / (2 * d);
} 

// case 02 : 's' is greater than 's1'
else{
    res = n1 + (s - s1) / m;
}

return res;

    }