Maximum of all subarrays of size k (Added a O(n) method) | GeeksforGeeks

https://www.geeksforgeeks.org/sliding-window-maximum-maximum-of-all-subarrays-of-size-k/
Given an array and an integer k, find the maximum for each and every contiguous subarray of size k.

Examples :

Input :
arr[] = {1, 2, 3, 1, 4, 5, 2, 3, 6}
k = 3
Output :
3 3 4 5 5 5 6

Input :
arr[] = {8, 5, 10, 7, 9, 4, 15, 12, 90, 13}
k = 4
Output :
10 10 10 15 15 90 90

(A O(n) method: use Dequeue)
We create a Dequeue, Qi of capacity k, that stores only useful elements of current window of k elements. An element is useful if it is in current window and is greater than all other elements on left side of it in current window. We process all array elements one by one and maintain Qi to contain useful elements of current window and these useful elements are maintained in sorted order. The element at front of the Qi is the largest and element at rear of Qi is the smallest of current window. Thanks to Aashish for suggesting this method.

Time Complexity: O(n). It seems more than O(n) at first look. If we take a closer look, we can observe that every element of array is added and removed at most once. So there are total 2n operations.

    // A Dequeue (Double ended queue) based method for printing maixmum element of

    // all subarrays of size k

    static void printMax(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

        Deque<Integer> Qi = new LinkedList<Integer>();

          

        /* 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.isEmpty() && arr[i] >= arr[Qi.peekLast()])

                Qi.removeLast();   // Remove from rear

              

            // Add new element at rear of queue

            Qi.addLast(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

            System.out.print(arr[Qi.peek()] + " ");

              

            // Remove the elements which are out of this window

            while((!Qi.isEmpty()) && Qi.peek() <= i-k)

                Qi.removeFirst();

              

            // Remove all elements smaller than the currently

            // being added element (remove useless elements)

            while((!Qi.isEmpty()) && arr[i] >= arr[Qi.peekLast()])

                Qi.removeLast();

              

  

            // Add current element at the rear of Qi

            Qi.addLast(i);

              

        }

          

        // Print the maximum element of last window

        System.out.print(arr[Qi.peek()]);

    }

https://www.geeksforgeeks.org/sum-minimum-maximum-elements-subarrays-size-k/

Given an array of both positive and negative integers, the task is to compute sum of minimum and maximum elements of all sub-array of size k.

Examples:

Input : arr[] = {2, 5, -1, 7, -3, -1, -2}  
        K = 4
Output : 18
Explanation : Subarrays of size 4 are : 
     {2, 5, -1, 7},   min + max = -1 + 7 = 6
     {5, -1, 7, -3},  min + max = -3 + 7 = 4      
     {-1, 7, -3, -1}, min + max = -3 + 7 = 4
     {7, -3, -1, -2}, min + max = -3 + 7 = 4   
     Sum of all min & max = 6 + 4 + 4 + 4 
                          = 18               

    // Returns sum of min and max element of all subarrays 

    // of size k 

    public static int SumOfKsubArray(int arr[] , int k) 

    { 

        int sum = 0;  // Initialize result 

    

        // The queue will store indexes of useful elements 

        // in every window 

        // In deque 'G' we maintain decreasing order of 

        // values from front to rear 

        // In deque 'S' we  maintain increasing order of 

        // values from front to rear 

        Deque<Integer> S=new LinkedList<>(),G=new LinkedList<>();

  

        // Process first window of size K 

        int i = 0; 

        for (i = 0; i < k; i++) 

        { 

            // Remove all previous greater elements 

            // that are useless. 

            while ( !S.isEmpty() && arr[S.peekLast()] >= arr[i]) 

                S.removeLast(); // Remove from rear 

    

            // Remove all previous smaller that are elements 

            // are useless. 

            while ( !G.isEmpty() && arr[G.peekLast()] <= arr[i]) 

                G.removeLast(); // Remove from rear 

    

            // Add current element at rear of both deque 

            G.addLast(i); 

            S.addLast(i); 

        } 

    

        // Process rest of the Array elements 

        for (  ; i < arr.length; i++ ) 

        { 

            // Element at the front of the deque 'G' & 'S' 

            // is the largest and smallest 

            // element of previous window respectively 

            sum += arr[S.peekFirst()] + arr[G.peekFirst()]; 

    

            // Remove all elements which are out of this 

            // window 

            while ( !S.isEmpty() && S.peekFirst() <= i - k) 

                S.removeFirst(); 

            while ( !G.isEmpty() && G.peekFirst() <= i - k) 

                G.removeFirst(); 

    

            // remove all previous greater element that are 

            // useless 

            while ( !S.isEmpty() && arr[S.peekLast()] >= arr[i]) 

                S.removeLast(); // Remove from rear 

    

            // remove all previous smaller that are elements 

            // are useless 

            while ( !G.isEmpty() && arr[G.peekLast()] <= arr[i]) 

                G.removeLast(); // Remove from rear 

    

            // Add current element at rear of both deque 

            G.addLast(i); 

            S.addLast(i); 

        } 

    

        // Sum of minimum and maximum element of last window 

        sum += arr[S.peekFirst()] + arr[G.peekFirst()]; 

    

        return sum; 

    } 

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