Difference between maximum and minimum average of all K-length contiguous subarrays
Last Updated :
23 Jul, 2025
Given an array arr[] of size N and an integer K, the task is to print the difference between the maximum and minimum average of the contiguous subarrays of length K.
Examples:
Input: arr[ ] = {3, 8, 9, 15}, K = 2
Output: 6.5
Explanation:
All subarrays of length 2 are {3, 8}, {8, 9}, {9, 15} and their averages are (3+8)/2 = 5.5, (8+9)/2 = 8.5, and (9+15)/2 = 12.0 respectively.
Therefore, the difference between the maximum(=12.0) and minimum(=5.5) is 12.0 -5.5 = 6.5.
Input: arr[] = {17, 6.2, 19, 3.4}, K = 3
Output: 4.533
Naive Approach: The simplest approach is to find the average of every contiguous subarray of size K, and then find the maximum and minimum of these values, and print their difference.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the difference between
// averages of the maximum and the minimum
// subarrays of length k
double Avgdifference(double arr[], int N, int K)
{
// Stores min and max sum
double min = 1000000, max = -1;
// Iterate through starting points
for (int i = 0; i <= N - K; i++) {
double sum = 0;
// Sum up next K elements
for (int j = 0; j < K; j++) {
sum += arr[i + j];
}
// Update max and min moving sum
if (min > sum)
min = sum;
if (max < sum)
max = sum;
}
// Return the difference between max
// and min average
return (max - min) / K;
}
// Driver Code
int main()
{
// Given Input
double arr[] = { 3, 8, 9, 15 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 2;
// Function Call
cout << Avgdifference(arr, N, K);
return 0;
}
// Java implementation of the above approach
import java.io.*;
class GFG
{
// Function to find the difference between
// averages of the maximum and the minimum
// subarrays of length k
static double Avgdifference(double arr[], int N, int K)
{
// Stores min and max sum
double min = 1000000, max = -1;
// Iterate through starting points
for (int i = 0; i <= N - K; i++) {
double sum = 0;
// Sum up next K elements
for (int j = 0; j < K; j++) {
sum += arr[i + j];
}
// Update max and min moving sum
if (min > sum)
min = sum;
if (max < sum)
max = sum;
}
// Return the difference between max
// and min average
return (max - min) / K;
}
// Driver Code
public static void main (String[] args)
{
// Given Input
double arr[] = { 3, 8, 9, 15 };
int N =arr.length;
int K = 2;
// Function Call
System.out.println( Avgdifference(arr, N, K));
}
}
// This code is contributed by Potta Lokesh
# Python program for the above approach
# Function to find the difference between
# averages of the maximum and the minimum
# subarrays of length k
def Avgdifference(arr, N, K):
# Stores min and max sum
min = 1000000;
max = -1;
# Iterate through starting points
for i in range(N - K + 1):
sum = 0;
# Sum up next K elements
for j in range(K):
sum += arr[i + j];
# Update max and min moving sum
if (min > sum):
min = sum;
if (max < sum):
max = sum;
# Return the difference between max
# and min average
return (max - min) / K;
# Driver Code
# Given Input
arr = [3, 8, 9, 15];
N = len(arr);
K = 2;
# Function Call
print(Avgdifference(arr, N, K));
# This code is contributed by _saurabh_jaiswal.
// C# program for the above approach
using System;
class GFG{
// Function to find the difference between
// averages of the maximum and the minimum
// subarrays of length k
static double Avgdifference(double []arr, int N, int K)
{
// Stores min and max sum
double min = 1000000, max = -1;
// Iterate through starting points
for(int i = 0; i <= N - K; i++)
{
double sum = 0;
// Sum up next K elements
for(int j = 0; j < K; j++)
{
sum += arr[i + j];
}
// Update max and min moving sum
if (min > sum)
min = sum;
if (max < sum)
max = sum;
}
// Return the difference between max
// and min average
return(max - min) / K;
}
// Driver Code
public static void Main (String[] args)
{
// Given Input
double []arr = { 3, 8, 9, 15 };
int N = arr.Length;
int K = 2;
// Function Call
Console.Write(Avgdifference(arr, N, K));
}
}
// This code is contributed by shivanisinghss2110
<script>
// JavaScript program for the above approach
// Function to find the difference between
// averages of the maximum and the minimum
// subarrays of length k
function Avgdifference(arr, N, K) {
// Stores min and max sum
let min = 1000000, max = -1;
// Iterate through starting points
for (let i = 0; i <= N - K; i++) {
let sum = 0;
// Sum up next K elements
for (let j = 0; j < K; j++) {
sum += arr[i + j];
}
// Update max and min moving sum
if (min > sum)
min = sum;
if (max < sum)
max = sum;
}
// Return the difference between max
// and min average
return (max - min) / K;
}
// Driver Code
// Given Input
let arr = [3, 8, 9, 15];
let N = arr.length;
let K = 2;
// Function Call
document.write(Avgdifference(arr, N, K));
</script>
Time Complexity: O(N*K)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized using the sliding window technique. Follow the steps below to solve the problem:
- Find the sum of subarrays over the range [0, K-1] and store it in a variable say sum.
- Initialize two variables, say max and min, to store the maximum and minimum sum of any subarray of size K.
- Iterate over the range [K, N-K+1] using the variable i and perform the following steps:
- Remove the element arr[i-K] and add the element arr[i] to the windows of size K. i.e Update sum to sum+arr[i]-arr[i-K].
- Update the min as the minimum of min and sum, and update max as the maximum of max and sum.
- Finally, after completing the above steps, print the answer as (max-min)/K.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the difference between
// the maximum and minimum subarrays of
// length K
double Avgdifference(double arr[], int N, int K)
{
// Stores the sum of subarray over the
// range [0, K]
double sum = 0;
// Iterate over the range [0, K]
for (int i = 0; i < K; i++)
sum += arr[i];
// Store min and max sum
double min = sum;
double max = sum;
// Iterate over the range [K, N-K]
for (int i = K; i <= N - K + 1; i++) {
// Increment sum by arr[i]-arr[i-K]
sum += arr[i] - arr[i - K];
// Update max and min moving sum
if (min > sum)
min = sum;
if (max < sum)
max = sum;
}
// Return difference between max and min
// average
return (max - min) / K;
}
// Driver Code
int main()
{
// Given Input
double arr[] = { 3, 8, 9, 15 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 2;
// Function Call
cout << Avgdifference(arr, N, K);
return 0;
}
// Java program for the above approach
import java.io.*;
class GFG{
// Function to find the difference between
// the maximum and minimum subarrays of
// length K
static double Avgdifference(double arr[], int N, int K)
{
// Stores the sum of subarray over the
// range [0, K]
double sum = 0;
// Iterate over the range [0, K]
for(int i = 0; i < K; i++)
sum += arr[i];
// Store min and max sum
double min = sum;
double max = sum;
// Iterate over the range [K, N-K]
for(int i = K; i <= N - K + 1; i++)
{
// Increment sum by arr[i]-arr[i-K]
sum += arr[i] - arr[i - K];
// Update max and min moving sum
if (min > sum)
min = sum;
if (max < sum)
max = sum;
}
// Return difference between max and min
// average
return(max - min) / K;
}
// Driver Code
public static void main (String[] args)
{
// Given Input
double arr[] = { 3, 8, 9, 15 };
int N = arr.length;
int K = 2;
// Function Call
System.out.println(Avgdifference(arr, N, K));
}
}
// This code is contributed by shivanisinghss2110
# python 3 program for the above approach
# Function to find the difference between
# the maximum and minimum subarrays of
# length K
def Avgdifference(arr, N, K):
# Stores the sum of subarray over the
# range [0, K]
sum = 0
# Iterate over the range [0, K]
for i in range(K):
sum += arr[i]
# Store min and max sum
min = sum
max = sum
# Iterate over the range [K, N-K]
for i in range(K,N - K + 2,1):
# Increment sum by arr[i]-arr[i-K]
sum += arr[i] - arr[i - K]
# Update max and min moving sum
if (min > sum):
min = sum
if (max < sum):
max = sum
# Return difference between max and min
# average
return (max - min) / K
# Driver Code
if __name__ == '__main__':
# Given Input
arr = [3, 8, 9, 15]
N = len(arr)
K = 2
# Function Call
print(Avgdifference(arr, N, K))
# This code is contributed by ipg2016107.
// C# program for the above approach
using System;
class GFG{
// Function to find the difference between
// the maximum and minimum subarrays of
// length K
static double Avgdifference(double []arr, int N, int K)
{
// Stores the sum of subarray over the
// range [0, K]
double sum = 0;
// Iterate over the range [0, K]
for(int i = 0; i < K; i++)
sum += arr[i];
// Store min and max sum
double min = sum;
double max = sum;
// Iterate over the range [K, N-K]
for(int i = K; i <= N - K + 1; i++)
{
// Increment sum by arr[i]-arr[i-K]
sum += arr[i] - arr[i - K];
// Update max and min moving sum
if (min > sum)
min = sum;
if (max < sum)
max = sum;
}
// Return difference between max and min
// average
return(max - min) / K;
}
// Driver Code
public static void Main (String[] args)
{
// Given Input
double []arr = { 3, 8, 9, 15 };
int N = arr.Length;
int K = 2;
// Function Call
Console.Write(Avgdifference(arr, N, K));
}
}
// This code is contributed by shivanisinghss2110
<script>
// JavaScript program for the above approach
// Function to find the difference between
// the maximum and minimum subarrays of
// length K
function Avgdifference(arr, N, K)
{
// Stores the sum of subarray over the
// range [0, K]
let sum = 0;
// Iterate over the range [0, K]
for(let i = 0; i < K; i++)
sum += arr[i];
// Store min and max sum
let min = sum;
let max = sum;
// Iterate over the range [K, N-K]
for(let i = K; i <= N - K + 1; i++)
{
// Increment sum by arr[i]-arr[i-K]
sum += arr[i] - arr[i - K];
// Update max and min moving sum
if (min > sum)
min = sum;
if (max < sum)
max = sum;
}
// Return difference between max and min
// average
return(max - min) / K;
}
// Driver Code
// Given Input
let arr = [ 3, 8, 9, 15 ];
let N = arr.length;
let K = 2;
// Function Call
document.write(Avgdifference(arr, N, K));
// This code is contributed by shivanisinghss2110
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
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