Sum of minimum and maximum elements of all subarrays of 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

This problem is mainly an extension of below problem.
Maximum of all subarrays of size k

Method 1 (Simple)
Run two loops to generate all subarrays of size k and find maximum and minimum values. Finally return sum of all maximum and minimum elements.

Time taken by this solution is O(nk).

Method 2 (Efficient using Dequeue)
The idea is to use Dequeue data structure and sliding window concept. We create two empty double ended queues of size k (‘S’ , ‘G’) that only store indexes of elements of current window that are not useless. An element is useless if it can not be maximum or minimum of next subarrays.

 a) In deque 'G', we maintain decreasing order of 
    values from front to rear
 b) In deque 'S', we maintain increasing order of 
    values from front to rear

1) First window size K
  1.1) For deque 'G', if current element is greater 
       than rear end element, we remove rear while 
       current is greater.
  1.2) For deque 'S', if current element is smaller 
       than rear end element, we just pop it while 
       current is smaller.
  1.3) insert current element in both deque 'G' 'S'

2) After step 1, front of 'G' contains maximum element
   of first window and front of 'S' contains minimum 
   element of first window. Remaining elements of G
   and S may store maximum/minimum for subsequent 
   windows.

3) After that we do traversal for rest array elements.
  3.1) Front element of deque 'G' is greatest and 'S' 
       is smallest element of previous window 
  3.2) Remove all elements which are out of this 
       window [remove element at front of queue ]
  3.3) Repeat steps 1.1 , 1.2 ,1.3 

4) Return sum of minimum and maximum element of all 
   sub-array size k.

Below is implementation of above idea

C++

// C++ program to find sum of all minimum and maximum 
// elements Of Sub-array Size k. 
#include<bits/stdc++.h> 
using namespace std; 
  
// Returns sum of min and max element of all subarrays 
// of size k 
int SumOfKsubArray(int arr[] , int n , 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< int > S(k), G(k); 
  
    // 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.empty()) && arr[S.back()] >= arr[i]) 
            S.pop_back(); // Remove from rear 
  
        // Remove all previous smaller that are elements 
        // are useless. 
        while ( (!G.empty()) && arr[G.back()] <= arr[i]) 
            G.pop_back(); // Remove from rear 
  
        // Add current element at rear of both deque 
        G.push_back(i); 
        S.push_back(i); 
    } 
  
    // Process rest of the Array elements 
    for (  ; i < n; i++ ) 
    { 
        // Element at the front of the deque 'G' & 'S' 
        // is the largest and smallest 
        // element of previous window respectively 
        sum += arr[S.front()] + arr[G.front()]; 
  
        // Remove all elements which are out of this 
        // window 
        while ( !S.empty() && S.front() <= i - k) 
            S.pop_front(); 
        while ( !G.empty() && G.front() <= i - k) 
            G.pop_front(); 
  
        // remove all previous greater element that are 
        // useless 
        while ( (!S.empty()) && arr[S.back()] >= arr[i]) 
            S.pop_back(); // Remove from rear 
  
        // remove all previous smaller that are elements 
        // are useless 
        while ( (!G.empty()) && arr[G.back()] <= arr[i]) 
            G.pop_back(); // Remove from rear 
  
        // Add current element at rear of both deque 
        G.push_back(i); 
        S.push_back(i); 
    } 
  
    // Sum of minimum and maximum element of last window 
    sum += arr[S.front()] + arr[G.front()]; 
  
    return sum; 
} 
  
// Driver program to test above functions 
int main() 
{ 
    int arr[] = {2, 5, -1, 7, -3, -1, -2} ; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    int k = 3; 
    cout << SumOfKsubArray(arr, n, k) ; 
    return 0; 
} 

Java

// Java program to find sum of all minimum and maximum  
// elements Of Sub-array Size k.  
import java.util.Deque; 
import java.util.LinkedList; 
public class Geeks { 
  
    // 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;  
    }  
  
    public static void main(String args[])  
    { 
        int arr[] = {2, 5, -1, 7, -3, -1, -2} ;  
        int k = 3;  
        System.out.println(SumOfKsubArray(arr, k)); 
    } 
} 
//This code is contributed by Gaurav Tiwari 

Python

# Python3 program to find Sum of all minimum and maximum  
# elements Of Sub-array Size k. 
from collections import deque 
  
# Returns Sum of min and max element of all subarrays 
# of size k 
def SumOfKsubArray(arr, n , k): 
  
    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 
    S = deque() 
    G = deque() 
  
  
    # Process first window of size K 
  
    for i in range(k): 
          
        # Remove all previous greater elements 
        # that are useless. 
        while ( len(S) > 0 and arr[S[-1]] >= arr[i]): 
            S.pop() # Remove from rear 
  
        # Remove all previous smaller that are elements 
        # are useless. 
        while ( len(G) > 0 and arr[G[-1]] <= arr[i]): 
            G.pop() # Remove from rear 
  
        # Add current element at rear of both deque 
        G.append(i) 
        S.append(i) 
  
    # Process rest of the Array elements 
    for i in range(k, n): 
          
        # Element at the front of the deque 'G' & 'S' 
        # is the largest and smallest 
        # element of previous window respectively 
        Sum += arr[S[0]] + arr[G[0]] 
  
        # Remove all elements which are out of this 
        # window 
        while ( len(S) > 0 and S[0] <= i - k): 
            S.popleft() 
        while ( len(G) > 0 and G[0] <= i - k): 
            G.popleft() 
  
        # remove all previous greater element that are 
        # useless 
        while ( len(S) > 0 and arr[S[-1]] >= arr[i]): 
            S.pop() # Remove from rear 
  
        # remove all previous smaller that are elements 
        # are useless 
        while ( len(G) > 0 and arr[G[-1]] <= arr[i]): 
            G.pop() # Remove from rear 
  
        # Add current element at rear of both deque 
        G.append(i) 
        S.append(i) 
  
    # Sum of minimum and maximum element of last window 
    Sum += arr[S[0]] + arr[G[0]] 
  
    return Sum
  
# Driver program to test above functions 
arr=[2, 5, -1, 7, -3, -1, -2] 
n = len(arr) 
k = 3
print(SumOfKsubArray(arr, n, k)) 
  
# This code is contributed by mohit kumar 

Output:

Time Complexity: O(n)

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