Maximum absolute difference between sum of subarrays of size K
Last Updated :
15 Jul, 2025
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
Brute Force Approach:
- Initialize a variable max_diff to zero.
- Loop through all pairs of starting indices i and j, where i ranges from 0 to N-K and j ranges from i+1 to N-K.
- For each pair of starting indices i and j, compute the sum of the subarray of size K starting at i, and the sum of the subarray of size K starting at j. To do this, we use a loop that adds up the K elements starting at i and j, respectively.
- Compute the absolute difference between the two sums computed in the previous step, and update the max_diff variable if this absolute difference is greater than the current value of max_diff.
- Return the value of max_diff.
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)
{
int max_diff = 0;
for (int i = 0; i <= N-K; i++) {
for (int j = i+1; j <= N-K; j++) {
int sum1 = 0, sum2 = 0;
for (int k = 0; k < K; k++) {
sum1 += arr[i+k];
sum2 += arr[j+k];
}
max_diff = max(max_diff, abs(sum1 - sum2));
}
}
return max_diff;
}
// 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;
}
import java.util.*;
public class Main {
// Return absolute difference
// between sum of all subarrays
// of size k
static int MaxAbsSumOfKsubArray(int[] arr, int K, int N) {
int max_diff = 0;
for (int i = 0; i <= N - K; i++) {
for (int j = i + 1; j <= N - K; j++) {
int sum1 = 0, sum2 = 0;
for (int k = 0; k < K; k++) {
sum1 += arr[i + k];
sum2 += arr[j + k];
}
max_diff = Math.max(max_diff, Math.abs(sum1 - sum2));
}
}
return max_diff;
}
// 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));
}
}
// This code is contributed by Prajwal Kandekar
def max_abs_sum_of_k_subarray(arr, K, N):
# Function to find the maximum absolute difference between the sums of two subarrays of length K
max_diff = 0
# Iterate through all possible starting indices of the first subarray
for i in range(N-K+1):
# Iterate through all possible starting indices of the second subarray
for j in range(i+1, N-K+1):
sum1 = 0
sum2 = 0
# Compute the sum of elements in the first subarray
for k in range(K):
sum1 += arr[i+k]
# Compute the sum of elements in the second subarray
for k in range(K):
sum2 += arr[j+k]
# Update the maximum difference if the absolute difference between the sums is greater
max_diff = max(max_diff, abs(sum1 - sum2))
return max_diff
# Driver code
arr = [-2, -3, 4, -1, -2, 1, 5, -3]
K = 3
N = len(arr)
print(max_abs_sum_of_k_subarray(arr, K, N))
// C# program to find the
// maximum absolute difference
// between the sum of all
// subarrays of size K
using System;
public static class GFG {
// Return absolute difference
// between sum of all subarrays
// of size k
static int MaxAbsSumOfKsubArray(int[] arr, int K, int N) {
int max_diff = 0;
for (int i = 0; i <= N - K; i++) {
for (int j = i + 1; j <= N - K; j++) {
int sum1 = 0, sum2 = 0;
for (int k = 0; k < K; k++) {
sum1 += arr[i + k];
sum2 += arr[j + k];
}
max_diff = Math.Max(max_diff, Math.Abs(sum1 - sum2));
}
}
return max_diff;
}
// Driver code
public static void Main() {
int[] arr = { -2, -3, 4, -1, -2, 1, 5, -3 };
int K = 3;
int N = arr.Length;
Console.WriteLine(MaxAbsSumOfKsubArray(arr, K, N));
}
}
// This code is contributed by Utkarsh Kumar
// Javascript program
// Return absolute difference
// between sum of all subarrays
// of size k
function maxAbsSumOfKsubArray(arr, K, N) {
let maxDiff = 0;
for (let i = 0; i <= N - K; i++) {
for (let j = i + 1; j <= N - K; j++) {
let sum1 = 0, sum2 = 0;
for (let k = 0; k < K; k++) {
sum1 += arr[i + k];
sum2 += arr[j + k];
}
maxDiff = Math.max(maxDiff, Math.abs(sum1 - sum2));
}
}
return maxDiff;
}
// Driver code
const arr = [-2, -3, 4, -1, -2, 1, 5, -3];
const K = 3;
const N = arr.length;
console.log(maxAbsSumOfKsubArray(arr, K, N));
Time Complexity: O(N^3)
Auxiliary Space: O(1)
Efficient Approach
The idea is to use Sliding Window Technique. Follow the steps below to solve the problem:
- Check if K is greater than N then return -1.
-
- 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.
- Calculate the sum of first K size subarray and update maxSum and minSum, decrement sum by arr[start] and increment start by 1.
- 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.
- 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 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 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# 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
<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>
Time Complexity: O (N)
Auxiliary Space: O (1)
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