# 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 c++ implementation of above idea

```// 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;
}
```

Output:

```16
```

Time Complexity: O(n)

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