Sum of minimum and maximum elements of all subarrays of size k.

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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   
     
     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 <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)
{
    // 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

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 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

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

<script>
 
// 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 , k)
{
    let 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
    let S = [];
    let G = [];
 
    // Process first window of size K
    let i = 0;
    for (i = 0; i < k; i++)
    {
 
    // Remove all previous greater elements
    // that are useless.
    while ( S.length != 0 && arr[S[S.length - 1]] >= arr[i])
        S.pop(); // Remove from rear
 
    // Remove all previous smaller that are elements
    // are useless.
    while ( G.length != 0 && arr[G[G.length - 1]] <= arr[i])
        G.pop(); // Remove from rear
 
    // Add current element at rear of both deque
    G.push(i);
    S.push(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.length != 0 && S[0] <= i - k)
        S.shift(0);
    while ( G.length != 0 && G[0] <= i - k)
        G.shift(0);
 
    // remove all previous greater element that are
    // useless
    while ( S.length != 0 && arr[S[S.length-1]] >= arr[i])
        S.pop(); // Remove from rear
 
    // remove all previous smaller that are elements
    // are useless
    while ( G.length != 0 && arr[G[G.length - 1]] <= arr[i])
        G.pop(); // Remove from rear
 
    // Add current element at rear of both deque
    G.push(i);
    S.push(i);
    }
 
    // Sum of minimum and maximum element of last window
    sum += arr[S[0]] + arr[G[0]];
    return sum;
}
 
// Driver code
 
    let arr = [2, 5, -1, 7, -3, -1, -2];
    let k = 3;
    document.write(SumOfKsubArray(arr, k));
 
// This code is contributed by _saurabh_jaiswal
 
</script>

Output

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

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