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# 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 = 4Output : 18Explanation : 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             Missing sub arrays -           {2, -1, 7, -3}     {2, 7, -3, -1}     {2, -3, -1, -2}     {5, 7, -3, -1}     {5, -3, -1, -2}     and few more -- why these were not considered??     Considering missing arrays result coming as 27          Sum of all min & max = 6 + 4 + 4 + 4                                = 18               `

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

Naive Approach: Run two loops to generate all subarrays and then choose all subarrays of size k and find maximum and minimum values. Finally, return the sum of all maximum and minimum elements.

Steps to implement-

• Initialize a variable sum with value 0 to store the final answer
• Run two loops to find all subarrays
• Simultaneously find the length of the subarray
• If there is any subarray of size k
• Then find its maximum and minimum element
• Then add that to the sum variable
• In the last print/return value stored in the sum variable

Code-

## C++

 `// C++ program to find sum of all minimum and maximum``// elements Of Sub-array Size k.``#include ``using` `namespace` `std;` `// Returns sum of min and max element of all subarrays``// of size k``int` `SumOfKsubArray(``int` `arr[], ``int` `N, ``int` `k)``{``    ``// To store final answer``    ``int` `sum = 0;` `    ``// Find all subarray``    ``for` `(``int` `i = 0; i < N; i++) {``        ``// To store length of subarray``        ``int` `length = 0;``        ``for` `(``int` `j = i; j < N; j++) {``            ``// Increment the length``            ``length++;` `            ``// When there is subarray of size k``            ``if` `(length == k) {``                ``// To store maximum and minimum element``                ``int` `maxi = INT_MIN;``                ``int` `mini = INT_MAX;` `                ``for` `(``int` `m = i; m <= j; m++) {``                    ``// Find maximum and minimum element``                    ``maxi = max(maxi, arr[m]);``                    ``mini = min(mini, arr[m]);``                ``}` `                ``// Add maximum and minimum element in sum``                ``sum += maxi + mini;``            ``}``        ``}``    ``}``    ``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.Arrays;` `class` `GFG {``    ``// Returns sum of min and max element of all subarrays``    ``// of size k``    ``static` `int` `SumOfKsubArray(``int``[] arr, ``int` `N, ``int` `k) {``        ``// To store the final answer``        ``int` `sum = ``0``;` `        ``// Find all subarrays``        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``// To store the length of the subarray``            ``int` `length = ``0``;``            ``for` `(``int` `j = i; j < N; j++) {``                ``// Increment the length``                ``length++;` `                ``// When there is a subarray of size k``                ``if` `(length == k) {``                    ``// To store the maximum and minimum element``                    ``int` `maxi = Integer.MIN_VALUE;``                    ``int` `mini = Integer.MAX_VALUE;` `                    ``for` `(``int` `m = i; m <= j; m++) {``                        ``// Find the maximum and minimum element``                        ``maxi = Math.max(maxi, arr[m]);``                        ``mini = Math.min(mini, arr[m]);``                    ``}` `                    ``// Add the maximum and minimum element to the sum``                    ``sum += maxi + mini;``                ``}``            ``}``        ``}``        ``return` `sum;``    ``}` `    ``// Driver program to test above functions``    ``public` `static` `void` `main(String[] args) {``        ``int``[] arr = {``2``, ``5``, -``1``, ``7``, -``3``, -``1``, -``2``};``        ``int` `N = arr.length;``        ``int` `k = ``3``;``        ``System.out.println(SumOfKsubArray(arr, N, k));``    ``}``}``//This code is contributed by Vishal Dhaygude`

## Python

 `# Returns sum of min and max element of all subarrays``# of size k``def` `sum_of_k_subarray(arr, N, k):``    ``# To store final answer``    ``sum` `=` `0` `    ``# Find all subarrays``    ``for` `i ``in` `range``(N):``        ``# To store length of subarray``        ``length ``=` `0``        ``for` `j ``in` `range``(i, N):``            ``# Increment the length``            ``length ``+``=` `1` `            ``# When there is a subarray of size k``            ``if` `length ``=``=` `k:``                ``# To store maximum and minimum element``                ``maxi ``=` `float``(``'-inf'``)``                ``mini ``=` `float``(``'inf'``)` `                ``for` `m ``in` `range``(i, j ``+` `1``):``                    ``# Find maximum and minimum element``                    ``maxi ``=` `max``(maxi, arr[m])``                    ``mini ``=` `min``(mini, arr[m])` `                ``# Add maximum and minimum element to sum``                ``sum` `+``=` `maxi ``+` `mini``    ``return` `sum` `# Driver program to test above function``def` `main():``    ``arr ``=` `[``2``, ``5``, ``-``1``, ``7``, ``-``3``, ``-``1``, ``-``2``]``    ``N ``=` `len``(arr)``    ``k ``=` `3``    ``print``(sum_of_k_subarray(arr, N, k))` `if` `__name__ ``=``=` `"__main__"``:``    ``main()`

## C#

 `using` `System;` `class` `Program {``    ``// Returns sum of min and max element of all subarrays``    ``// of size k``    ``static` `int` `SumOfKSubArray(``int``[] arr, ``int` `N, ``int` `k)``    ``{``        ``// To store the final answer``        ``int` `sum = 0;` `        ``// Find all subarrays``        ``for` `(``int` `i = 0; i < N; i++) {``            ``// To store the length of subarray``            ``int` `length = 0;``            ``for` `(``int` `j = i; j < N; j++) {``                ``// Increment the length``                ``length++;` `                ``// When there is a subarray of size k``                ``if` `(length == k) {``                    ``// To store the maximum and minimum``                    ``// element``                    ``int` `maxi = ``int``.MinValue;``                    ``int` `mini = ``int``.MaxValue;` `                    ``for` `(``int` `m = i; m <= j; m++) {``                        ``// Find maximum and minimum element``                        ``maxi = Math.Max(maxi, arr[m]);``                        ``mini = Math.Min(mini, arr[m]);``                    ``}` `                    ``// Add maximum and minimum element in``                    ``// sum``                    ``sum += maxi + mini;``                ``}``            ``}``        ``}``        ``return` `sum;``    ``}` `    ``// Driver program to test above functions``    ``static` `void` `Main()``    ``{``        ``int``[] arr = { 2, 5, -1, 7, -3, -1, -2 };``        ``int` `N = arr.Length;``        ``int` `k = 3;``        ``Console.WriteLine(SumOfKSubArray(arr, N, k));``    ``}``}`

## Javascript

 `// JavaScript program to find sum of all minimum and maximum``// elements of sub-array size k.` `// Returns sum of min and max element of all subarrays``// of size k``function` `SumOfKsubArray(arr, N, k) {``    ``// To store final answer``    ``let sum = 0;` `    ``// Find all subarray``    ``for` `(let i = 0; i < N; i++) {``        ``// To store length of subarray``        ``let length = 0;``        ``for` `(let j = i; j < N; j++) {``            ``// Increment the length``            ``length++;` `            ``// When there is subarray of size k``            ``if` `(length === k) {``                ``// To store maximum and minimum element``                ``let maxi = Number.MIN_SAFE_INTEGER;``                ``let mini = Number.MAX_SAFE_INTEGER;` `                ``for` `(let m = i; m <= j; m++) {``                    ``// Find maximum and minimum element``                    ``maxi = Math.max(maxi, arr[m]);``                    ``mini = Math.min(mini, arr[m]);``                ``}` `                ``// Add maximum and minimum element in sum``                ``sum += maxi + mini;``            ``}``        ``}``    ``}``    ``return` `sum;``}` `// Driver program to test above function``const arr = [2, 5, -1, 7, -3, -1, -2];``const N = arr.length;``const k = 3;``console.log(SumOfKsubArray(arr, N, k));`

Output

```14

```

Time Complexity: O(N2*k), because two loops to find all subarray and one loop to find the maximum and minimum elements in the subarray of size k
Auxiliary Space: O(1), because no extra space has been used

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 indices 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 rear1) 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``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 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`

## C#

 `// C# program to find sum of all minimum and maximum``// elements Of Sub-array Size k.``using` `System;``using` `System.Collections.Generic;``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``    ``List<``int``> S = ``new` `List<``int``>();``    ``List<``int``> G = ``new` `List<``int``>();` `    ``// 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.Count != 0 && arr[S[S.Count - 1]] >= arr[i])``        ``S.RemoveAt(S.Count - 1); ``// Remove from rear` `      ``// Remove all previous smaller that are elements``      ``// are useless.``      ``while` `( G.Count != 0 && arr[G[G.Count - 1]] <= arr[i])``        ``G.RemoveAt(G.Count - 1); ``// Remove from rear` `      ``// Add current element at rear of both deque``      ``G.Add(i);``      ``S.Add(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[0]] + arr[G[0]];` `      ``// Remove all elements which are out of this``      ``// window``      ``while` `( S.Count != 0 && S[0] <= i - k)``        ``S.RemoveAt(0);``      ``while` `( G.Count != 0 && G[0] <= i - k)``        ``G.RemoveAt(0);` `      ``// remove all previous greater element that are``      ``// useless``      ``while` `( S.Count != 0 && arr[S[S.Count-1]] >= arr[i])``        ``S.RemoveAt(S.Count - 1 ); ``// Remove from rear` `      ``// remove all previous smaller that are elements``      ``// are useless``      ``while` `( G.Count != 0 && arr[G[G.Count - 1]] <= arr[i])``        ``G.RemoveAt(G.Count - 1); ``// Remove from rear` `      ``// Add current element at rear of both deque``      ``G.Add(i);``      ``S.Add(i);``    ``}` `    ``// Sum of minimum and maximum element of last window``    ``sum += arr[S[0]] + arr[G[0]];  ``    ``return` `sum;``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main(String []args)``  ``{``    ``int` `[]arr = {2, 5, -1, 7, -3, -1, -2} ;``    ``int` `k = 3;``    ``Console.WriteLine(SumOfKsubArray(arr, k));``  ``}``}` `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output

```14

```

Time Complexity: O(n)
Auxiliary Space: O(k)

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