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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Chocolate Feast
  5. Discussions

Chocolate Feast

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1278 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my easy c++ solution , you can watch the explanation here : https://youtu.be/RZti_qPbiAA

    int chocolateFeast(int n, int c, int m) {
    int result = 0, wrappers = 0;
    while( n >= c){
        result ++;
        wrappers ++;
        n -= c;
        if(wrappers == m){
            result ++;
            wrappers = 1;
        }
    }
    return result;
}

  • somaditya
     + 0 comments

    Java: b for bars, w for wrappers

     public static int chocolateFeast(int n, int c, int m) {
        int b = n/c;
        int w = b;
        
        while (w >= m) {
            b += w/m;
            w = w%m + w/m;
        }
        
        return b;
    }

  • shuvi_pasko
     + 0 comments

    python solution

    recursive

    def recursiveChocolateFeast(wrapper,m):
    if wrapper<m:
        return 0
    return wrapper//m +recursiveChocolateFeast(wrapper//m+wrapper%m,m)
    
def chocolateFeast(n, c, m):
    return n//c +recursiveChocolateFeast(n//c,m)

    non recursive

    def chocolateFeast(n, c, m):
    # Write your code here
    res =n//c
    wrapper =res
    while wrapper>=m:
        temp = wrapper%m
        wrapper //=m
				res +=wrapper
        wrapper +=temp
		return res

    return res

  • khuatducminh_t67
     + 0 comments

    My solution in Python 3:

    def chocolateFeast(n, c, m):
    # Write your code here
    #n is the number of money Bobby has, he uses once
    #Bobby takes full advantage of promotion, hence, wrappers are used many times
        
    #Steps:
    '''
        1. Calculate intial bars, also intial wrappers.
        2. Create a loop, until Bobby cant trade, that updates bars and new wrappers (remain ones + traded ones)
        3. Return result
            
    '''
    
    bars: int = n / c 
    remain_wrappers: int = bars
    
    while remain_wrappers >= m:
        trade: int = remain_wrappers // m
        remain_wrappers = remain_wrappers % m + trade
        
        bars += trade
        
    return int(bars)

  • khuatducminh_t67
     + 0 comments

    My solution in Java 8:

    public static int chocolateFeast(int n, int c, int m) {
    // Write your code here
        //n is the number of money Bobby has, he uses once
        //Bobby takes full advantage of promotion, hence, wrappers are used many times
        
        //Steps:
        /*
            1. Calculate intial bars, also intial wrappers.
            2. Create a loop, until Bobby cant trade, that updates bars and new wrappers (remain ones + traded ones)
            3. Return result
            
        */
        
        int bars = n / c; //Initial bars
        int remainWrapper = bars; //Inital wrappers 
        
        while(remainWrapper >= m){
            //While he can trade
            int trade = remainWrapper / m;
            remainWrapper = remainWrapper - trade * m + trade;
            //remainWrapper = remainWrapper % m + trade;
            
            bars += trade;
        }
        
        return bars;
    }