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

Halloween Sale

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667 Discussions

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  • alban_tyrex
     + 0 comments

    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
}

  • aksaini2282003
     + 0 comments

    This is my approach

    def howManyGames(p, d, m, s): 
    n_games=0 

    while(s>=p):
            s-=p 
            n_games+=1
            p=max(p-d,m)
            
    return n_games

    `

  • patrickgao2020
     + 0 comments

    Here is my Python solution!

    def howManyGames(p, d, m, s):
    games = 0
    while True:
        if p > s:
            return games
        s -= p
        p = p - d if p - d >= m else m
        games += 1

  • kunalkawate424
     + 0 comments

    Solution From my side

    def howManyGames(p, d, m, s):
    results = []
    results.append(p)
    while(sum(results)<=s):
        if(p>m and (p - d)>m):
            p = p - d
            results.append(p)
        else:
            results.append(m)
    return len(results)-1

  • pub_ashishk28
     + 0 comments
    def howManyGames(p, d, m, s):
    # Return the number of games you can buy
     
    count = 0 
    i = p
    
    while s - i >= 0:
        count += 1
        s -= i
        i = max(m, i - d )
                   
    return count