Mark and Toys Discussions | Algorithms | HackerRank
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  3. Greedy
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Mark and Toys

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1033 Discussions

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  • 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;
}

  • air_mvp_jordan
     + 0 comments

    Using c#

    public static int maximumToys(List<int> prices, int k)
{
  int n = prices.Count;
  int toyCount = 0;
  int totalSpent = 0;

  prices.Sort();

  // Buy toys while the budget allows
  for (int i = 0; i < n; i++)
  {
    if (totalSpent + prices[i] <= k)
    {
      totalSpent += prices[i];
      toyCount++;
    }
    else
    {
      // If we can't afford this toy, stop the process
      break;
    }
  }
}

  • konthamshravani1
     + 0 comments

    Java Solution

    public static int maximumToys(List prices, int k) { // Write your code here Collections.sort(prices); int toys=0; /for(int i=0;i/

        int i=0;
    while(prices.get(i)<k && i<prices.size()){
        k=k-prices.get(i);
        toys=toys+1;
        i++;
    }
    return toys;

}

  • paschosachille
     + 0 comments

    Straigthforward python solution:

    def maximumToys(prices, k):

    sorted_prices = sorted(prices)
cost = 0    

for i in range(len(prices)):
    cost += sorted_prices[i]
    if cost > k:
        return i

  • herlianaNoviar
     + 0 comments

    PHP

    function maximumToys($prices, $k) {
    // Write your code here
    sort ($prices);
    $hasil   = 0;
    $counter = 0;
    
    for ($i=0; $i < count($prices); $i++) {
        if (($hasil+$prices[$i]) <= $k) {
            $hasil+=$prices[$i];
            $counter++;
        }
        else {
            break;
        }
    }
    return $counter;
}