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  1. Prepare
  2. C++
  3. STL
  4. Deque-STL
  5. Discussions

Deque-STL

  • mario135
     + 0 comments

    You need to find the max of the K values. Then while you do not will pop the max value, and not arrive the end of queue, continue comparing to the next value, printing the max and pop_front of queue. If the value to pop is the max value, then start again and find the max of the K values.

    This is not optimized and if a lot of max values are repeated the complexity will increment. But this code pass all the tests in time.

    void printKMax(int arr[], int n, int k){
	//Write your code here.
    deque<int> fila;
    deque<int>::iterator itr;
    int max = 0;
    
    for (int i = 0; i<n; i++)
    {
        fila.push_back(arr[i]);
    }
    
    while(fila.size() >= k)
    {
        itr = fila.begin();
        for(int i = 0; i<k ; i++)
        {
            if(*itr > max)
                max = *itr;
            itr++;
        }
        cout << max << " ";
        while ((fila.front() != max) && (itr != fila.end()))
        {
            if (max < *itr)
                max = *itr;
            cout << max << " ";
            itr++;
            fila.pop_front();
        }
        fila.pop_front();
        max = 0;
    }
    cout << endl;
}