Mark and Toys Discussions | Algorithms | HackerRank
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Mark and Toys

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1044 Discussions

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  • ai_13310411
     + 0 comments

    Java

    	Collections.sort(prices);
        int sum = 0;
        int count = 0;
        for(Integer i : prices){
            sum += i;
            if(sum > k){
                break;
            } else{
                count++;
            }
        }
        return count;

  • lylia_luong
     + 0 comments
    def maximumToys(prices, k):
    # Write your code here
    prices.sort()
    current=0
    toys=0
    for toy in prices:
        if toy+current<=k:
            current+=toy 
            toys+=1
    return toys

  • ahmet_alp77
     + 0 comments

    Java:

    public static int maximumToys(List<Integer> prices, int k) {
    prices.sort(Comparator.naturalOrder());
    int i = 0;
    int totalSpend = 0;
    for (;i < prices.size() && totalSpend < k; ++i) {
        totalSpend += prices.get(i);
    }
    return i-1;
}

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you also have a vidéo explanation here : https://youtu.be/DFQO52aB-qI

    int maximumToys(vector<int> prices, int k) {
    sort(prices.begin(), prices.end());
    int i = 0;
    while(i < prices.size() && k >= prices[i]){
        k -= prices[i];
        i++;
    }
    return i;
}

  • Phortran
     + 0 comments

    I've added a little optimisation to filter out elements above the budget, that we don't need spending time ordering.

    public static int maximumToys(List<Integer> prices, int k) {
        var sorted = prices.stream()
                           .filter(p -> p <= k)
                           .sorted()
                           .collect(Collectors.toList());
        
        var i = 0;
        var budget = k;
        
        while(i < sorted.size() && budget >= sorted.get(i)) {
            budget -= sorted.get(i);
            ++i;
        }
        
        return i;
    }