Maximum absolute difference between sum of subarrays of size K

Given an array arr[] of size N and an integer K, the task is to find maximum absolute difference between the sum of subarrays of size K.
Examples :

Input: arr[] = {-2, -3, 4, -1, -2, 1, 5, -3}, K = 3 
Output: 6 
Explanation:: 
Sum of subarray (-2, -3, 4) = -1 
Sum of subarray (-3, 4, -1) = 0 
Sum of subarray (4, -1, -2) = 1 
Sum of subarray (-1, -2, 1) = -2 
Sum of subarray (-2, 1, 5) = 4 
Sum of subarray (1, 5, -3) = 3 
So maximum absolute difference between sum of subarray of size 3 is is (4 – (-2)) = 6.
Input: arr [ ] = {2, 5, -1, 7, -3, -1, -2}, K = 4 
Output: 12 

Naive Approach
The simplest approach is to generate all subarrays of size K and find the minimum sum and maximum sum among them. Finally, return the absolute difference between the maximum and minimum sums. 
Time Complexity: O( N²)
Auxiliary Space: O (1)
Efficient Approach
The idea is to use Sliding Window Technique. Follow the steps below to solve the problem:

1. Check if K is greater than N then return -1.
2. • maxSum : Store maximum sum of K size subarray.
   • minSum : Store minimum sum of K size subarray.
   • sum : Store current sum of K size subarray.
   • start : Remove left most element which is no longer part of K size subarray.
3. Calculate the sum of first K size subarray and update maxSum and minSum, decrement sum by arr[start] and  increment start by 1.
4. Traverse arr from K to N and do the following operations: 
   • Increment sum by arr[i].
   • Update maxSum and minSum.
   • Decrement sum by arr[start].
   • Increment start by 1.
5. Return absolute difference between maxSum and minSum.

Below is the implementation of the above approach :

C++

// C++ program to find the
// maximum absolute difference
// between the sum of all
// subarrays of size K
#include <bits/stdc++.h>
using namespace std;

// Return absolute difference
// between sum of all subarrays
// of size k
int MaxAbsSumOfKsubArray(int arr[],
                         int K, int N)
{

    // Stores maximum sum of
    // all K size subarrays
    int maxSum = INT_MIN;

    // Stores minimum sum of
    // all K size subarray
    int minSum = INT_MAX;

    // Stores the sum of current
    // subarray of size K
    int sum = 0;

    // Starting index of the
    // current subarray
    int start = 0;

    int i = 0;

    if (N < K)
        return -1;

    // Calculate the sum of
    // first K elements
    while (i < K)
    {
        sum += arr[i];
        i++;
    }

    // Update maxSum and minSum
    maxSum = max(maxSum, sum);
    minSum = min(minSum, sum);

    // Decrement sum by arr[start]
    // and increment start by 1
    sum -= arr[start++];

    // Traverse arr for the
    // remaining subarrays
    while (i < N)
    {

        // Increment sum by arr[i]
        sum += arr[i];

        // Increment i
        i++;

        // Update maxSum and minSum
        maxSum = max(maxSum, sum);
        minSum = min(minSum, sum);

        // Decrement sum by arr[start]
        // and increment start by 1
        sum -= arr[start++];
    }

    // Return absolute difference
    // between maxSum and minSum
    return abs(maxSum - minSum);
}

// Driver code
int main()
{
    int arr[] = { -2, -3, 4, -1,
                  -2, 1, 5, -3 };
    int K = 3;
    int N = sizeof(arr) / sizeof(arr[0]);

    cout << MaxAbsSumOfKsubArray(arr, K, N)
         << endl;

    return 0;
}

// This code is contributed by divyeshrabadiya07

Java

// Java program to find the
// maximum absolute difference
// between the sum of all
// subarrays of size K

import java.util.*;

class GFG {

    // Return absolute difference
    // between sum of all subarrays
    // of size k
    static int MaxAbsSumOfKsubArray(
        int[] arr,
        int K, int N)
    {
        // Stores maximum sum of
        // all K size subarrays
        int maxSum = Integer.MIN_VALUE;

        // Stores minimum sum of
        // all K size subarray
        int minSum = Integer.MAX_VALUE;

        // Stores the sum of current
        // subarray of size K
        int sum = 0;

        // Starting index of the
        // current subarray
        int start = 0;

        int i = 0;

        if (N < K)
            return -1;

        // Calculate the sum of
        // first K elements
        while (i < K) {
            sum += arr[i];
            i++;
        }

        // Update maxSum and minSum
        maxSum = Math.max(maxSum, sum);
        minSum = Math.min(minSum, sum);

        // Decrement sum by arr[start]
        // and increment start by 1
        sum -= arr[start++];

        // Traverse arr for the
        // remaining subarrays
        while (i < N) {

            // Increment sum by arr[i]
            sum += arr[i];

            // Increment i
            i++;

            // Update maxSum and minSum
            maxSum = Math.max(maxSum, sum);
            minSum = Math.min(minSum, sum);

            // Decrement sum by arr[start]
            // and increment start by 1
            sum -= arr[start++];
        }

        // Return absolute difference
        // between maxSum and minSum
        return Math.abs(maxSum - minSum);
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { -2, -3, 4, -1,
                      -2, 1, 5, -3 };
        int K = 3;
        int N = arr.length;
        System.out.println(
            MaxAbsSumOfKsubArray(
                arr, K, N));
    }
}

Python3

# Python3 program to find the
# maximum absolute difference
# between the sum of all
# subarrays of size K
import sys

# Return absolute difference
# between sum of all subarrays
# of size k
def MaxAbsSumOfKsubArray(arr, K, N):

    # Stores maximum sum of
    # all K size subarrays
    maxSum = - sys.maxsize - 1

    # Stores minimum sum of
    # all K size subarray
    minSum = sys.maxsize

    # Stores the sum of current
    # subarray of size K
    sum = 0

    # Starting index of the
    # current subarray
    start = 0

    i = 0

    if (N < K):
        return -1

    # Calculate the sum of
    # first K elements
    while (i < K):
        sum += arr[i]
        i += 1

    # Update maxSum and minSum
    maxSum = max(maxSum, sum)
    minSum = min(minSum, sum)

    # Decrement sum by arr[start]
    # and increment start by 1
    sum -= arr[start]
    start += 1

    # Traverse arr for the
    # remaining subarrays
    while (i < N):

        # Increment sum by arr[i]
        sum += arr[i]

        # Increment i
        i += 1

        # Update maxSum and minSum
        maxSum = max(maxSum, sum)
        minSum = min(minSum, sum)

        # Decrement sum by arr[start]
        # and increment start by 1
        sum -= arr[start]
        start += 1

    # Return absolute difference
    # between maxSum and minSum
    return abs(maxSum - minSum)

# Driver code
arr = [ -2, -3, 4, -1,
        -2, 1, 5, -3 ]
K = 3
N = len(arr)

print(MaxAbsSumOfKsubArray(arr, K, N))

# This code is contributed by sanjoy_62

C#

// C# program to find the
// maximum absolute difference
// between the sum of all
// subarrays of size K
using System;
class GFG{

// Return absolute difference
// between sum of all subarrays
// of size k
static int MaxAbsSumOfKsubArray(
       int[] arr,
       int K, int N)
{
    // Stores maximum sum of
    // all K size subarrays
    int MaxSum = Int32.MinValue;

    // Stores minimum sum of
    // all K size subarray
    int MinSum = Int32.MaxValue;

    // Stores the sum of current
    // subarray of size K
    int sum = 0;

    // Starting index of the
    // current subarray
    int start = 0;

    int i = 0;

    if (N < K)
        return -1;

    // Calculate the sum of
    // first K elements
    while (i < K)
    {
        sum += arr[i];
        i++;
    }

    // Update maxSum and minSum
    MaxSum = Math.Max(MaxSum, sum);
    MinSum = Math.Min(MinSum, sum);

    // Decrement sum by arr[start]
    // and increment start by 1
    sum -= arr[start++];

    // Traverse arr for the
    // remaining subarrays
    while (i < N)
    {

        // Increment sum by arr[i]
        sum += arr[i];

        // Increment i
        i++;

        // Update maxSum and minSum
        MaxSum = Math.Max(MaxSum, sum);
        MinSum = Math.Min(MinSum, sum);

        // Decrement sum by arr[start]
        // and increment start by 1
        sum -= arr[start++];
    }

    // Return absolute difference
    // between maxSum and minSum
    return Math.Abs(MaxSum - MinSum);
}

// Driver code
public static void Main(String[] args)
{
    int[] arr = { -2, -3, 4, -1,
                  -2, 1, 5, -3 };
    int K = 3;
    int N = arr.Length;
    Console.Write(MaxAbsSumOfKsubArray(arr, K, N));
}
}

// This code is contributed
// by shivanisinghss2110

Javascript

<script>

    // Javascript program to find the
    // maximum absolute difference
    // between the sum of all
    // subarrays of size K

    // Return absolute difference
    // between sum of all subarrays
    // of size k
    function MaxAbsSumOfKsubArray(arr, K, N)
    {

        // Stores maximum sum of
        // all K size subarrays
        let maxSum = Number.MIN_VALUE;

        // Stores minimum sum of
        // all K size subarray
        let minSum = Number.MAX_VALUE;

        // Stores the sum of current
        // subarray of size K
        let sum = 0;

        // Starting index of the
        // current subarray
        let start = 0;

        let i = 0;

        if (N < K)
            return -1;

        // Calculate the sum of
        // first K elements
        while (i < K)
        {
            sum += arr[i];
            i++;
        }

        // Update maxSum and minSum
        maxSum = Math.max(maxSum, sum);
        minSum = Math.min(minSum, sum);

        // Decrement sum by arr[start]
        // and increment start by 1
        sum -= arr[start++];

        // Traverse arr for the
        // remaining subarrays
        while (i < N)
        {

            // Increment sum by arr[i]
            sum += arr[i];

            // Increment i
            i++;

            // Update maxSum and minSum
            maxSum = Math.max(maxSum, sum);
            minSum = Math.min(minSum, sum);

            // Decrement sum by arr[start]
            // and increment start by 1
            sum -= arr[start++];
        }

        // Return absolute difference
        // between maxSum and minSum
        return Math.abs(maxSum - minSum);
    }

    let arr = [ -2, -3, 4, -1, -2, 1, 5, -3 ];
    let K = 3;
    let N = arr.length;

    document.write(MaxAbsSumOfKsubArray(arr, K, N));

</script>

Output:

6

Time Complexity: O (N)
Auxiliary Space: O (1)