In this HackerRank Deque-STL problem in c++ programming you have Given a set of arrays of size N and an integer K, you have to find the maximum integer for each and every contiguous subarray of size K for each of the given arrays.


HackerRank Deque-STL solution in c++ programming


HackerRank Deque STL problem solution in c++ programming.

#include <iostream>
#include <deque> 
using namespace std;

void printKMax(int arr[], int n, int k){
	//Write your code here.
    deque<int> dq;
    
    for (int i=0; i<n; i++){
        
        // base case for first element
        if (dq.empty()){
            dq.push_back(i);
        }
        
        // remove elements outside the current window
        if (dq.front() <= (i - k)){
            dq.pop_front();
        }
        
        // move max element to the front
        while (!dq.empty() && arr[i] >= arr[dq.back()]){
            dq.pop_back();
        }
        
        dq.push_back(i);
        
        // print out only when the first window is completed
        if (i >= (k - 1)){
            cout << arr[dq.front()] << " ";
        }    
    }
    cout << endl;
}

int main(){
  
	int t;
	cin >> t;
	while(t>0) {
		int n,k;
    	cin >> n >> k;
    	int i;
    	int arr[n];
    	for(i=0;i<n;i++)
      		cin >> arr[i];
    	printKMax(arr, n, k);
    	t--;
  	}
  	return 0;
}


Second solution

#include <iostream>
#include <deque>

using namespace std;

// A Dequeue (Double ended queue) based method for printing maixmum element of
// all subarrays of size k
void printKMax(int arr[], int n, int k)
{
    // Create a Double Ended Queue, Qi that will store indexes of array elements
    // The queue will store indexes of useful elements in every window and it will
    // maintain decreasing order of values from front to rear in Qi, i.e., 
    // arr[Qi.front[]] to arr[Qi.rear()] are sorted in decreasing order
    std::deque<int>  Qi(k);

    /* Process first k (or first window) elements of array */
    int i;
    for (i = 0; i < k; ++i)
    {
        // For very element, the previous smaller elements are useless so
        // remove them from Qi
        while ( (!Qi.empty()) && arr[i] >= arr[Qi.back()])
            Qi.pop_back();  // Remove from rear

        // Add new element at rear of queue
        Qi.push_back(i);
    }

    // Process rest of the elements, i.e., from arr[k] to arr[n-1]
    for ( ; i < n; ++i)
    {
        // The element at the front of the queue is the largest element of
        // previous window, so print it
        cout << arr[Qi.front()] << " ";

        // Remove the elements which are out of this window
        while ( (!Qi.empty()) && Qi.front() <= i - k)
            Qi.pop_front();  // Remove from front of queue

        // Remove all elements smaller than the currently
        // being added element (remove useless elements)
        while ( (!Qi.empty()) && arr[i] >= arr[Qi.back()])
            Qi.pop_back();

         // Add current element at the rear of Qi
        Qi.push_back(i);
    }

    // Print the maximum element of last window
    cout << arr[Qi.front()] << endl;
}

// Driver program to test above functions
int main()
{
    int t;
    cin >> t;
    while(t>0) {
    int n,k;
    cin >> n >> k;
    int i;
    int arr[n];
    for(i=0;i<n;i++)
    cin >> arr[i];
    printKMax(arr, n, k);
    t--;
    }
    return 0;
}