Mark and Toys Discussions | Algorithms | HackerRank
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Mark and Toys

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1042 Discussions

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  • 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;
    }

  • emhhacker
     + 0 comments

    My Java solution with o(n log n) time complexity and o(1) space complexity:

    public static int maximumToys(List<Integer> prices, int k) {
        // sort array ascending
        Collections.sort(prices);
        
        // subtract price from k until k - price < 0
        int maxToys = 0;
        int i = 0;
        while(k - prices.get(i) >= 0){
            maxToys++;
            k -= prices.get(i);
            i++;
        }
        return maxToys;
    }

  • akash_prajapati6
     + 0 comments
    def maximumToys(prices, k):
    prices.sort()
    for i in range(len(prices)+1):
        if(sum(prices[0:i])<k):
            pass
        elif(sum(prices[0:i])>k):
            return i-1
        elif(sum(prices[0:i])==k):
            return i