# Find maximum average subarray of k length

• Difficulty Level : Easy
• Last Updated : 28 Jun, 2022

Given an array with positive and negative numbers, find the maximum average subarray of the given length.

Example:

Input:  arr[] = {1, 12, -5, -6, 50, 3}, k = 4
Output: Maximum average subarray of length 4 begins
at index 1.
Maximum average is (12 - 5 - 6 + 50)/4 = 51/4

A Simple Solution is to run two loops. The outer loop picks starting point, and the inner loop goes to length ‘k’ from the starting point and computes the average of elements.

Time Complexity: O(n*k), as we are using nested loops to traverse n*k times.
Auxiliary Space: O(1), as we are not using any extra space.

A Better Solution is to create an auxiliary array of size n. Store cumulative sum of elements in this array. Let the array be csum[]. csum[i] stores sum of elements from arr[0] to arr[i]. Once we have the csum[] array with us, we can compute the sum between two indexes in O(1) time.
Below is the implementation of this idea. One observation is, that a subarray of a given length has a maximum average if it has a maximum sum. So we can avoid floating-point arithmetic by just comparing sums.

## C++

 // C++ program to find maximum average subarray// of given length.#includeusing namespace std; // Returns beginning index of maximum average// subarray of length 'k'int findMaxAverage(int arr[], int n, int k){    // Check if 'k' is valid    if (k > n)        return -1;     // Create and fill array to store cumulative    // sum. csum[i] stores sum of arr[0] to arr[i]    int *csum = new int[n];    csum[0] = arr[0];    for (int i=1; i max_sum)        {            max_sum = curr_sum;            max_end = i;        }    }     delete [] csum; // To avoid memory leak     // Return starting index    return max_end - k + 1;} // Driver programint main(){    int arr[] = {1, 12, -5, -6, 50, 3};    int k = 4;    int n = sizeof(arr)/sizeof(arr[0]);    cout << "The maximum average subarray of "         "length "<< k << " begins at index "         << findMaxAverage(arr, n, k);    return 0;}

## Java

 // Java program to find maximum average// subarray of given length.import java .io.*; class GFG {     // Returns beginning index    // of maximum average    // subarray of length 'k'    static int findMaxAverage(int []arr,                           int n, int k)    {                 // Check if 'k' is valid        if (k > n)            return -1;             // Create and fill array        // to store cumulative        // sum. csum[i] stores        // sum of arr[0] to arr[i]        int []csum = new int[n];                 csum[0] = arr[0];        for (int i = 1; i < n; i++)        csum[i] = csum[i - 1] + arr[i];             // Initialize max_sm as        // sum of first subarray        int max_sum = csum[k - 1],                    max_end = k - 1;             // Find sum of other        // subarrays and update        // max_sum if required.        for (int i = k; i < n; i++)        {            int curr_sum = csum[i] -                    csum[i - k];            if (curr_sum > max_sum)            {                max_sum = curr_sum;                max_end = i;            }        }             // To avoid memory leak        //delete [] csum;                 // Return starting index        return max_end - k + 1;    }     // Driver Code    static public void main (String[] args)    {        int []arr = {1, 12, -5, -6, 50, 3};        int k = 4;        int n = arr.length;                 System.out.println("The maximum "          + "average subarray of length "                + k + " begins at index "            + findMaxAverage(arr, n, k));    }} // This code is contributed by anuj_67.

## Python3

 # Python program to find maximum average subarray# of given length. # Returns beginning index of maximum average# subarray of length 'k'def findMaxAverage(arr, n, k):    # Check if 'k' is valid    if k > n:        return -1     # Create and fill array to store cumulative    # sum. csum[i] stores sum of arr[0] to arr[i]    csum = [0]*n    csum[0] = arr[0]    for i in range(1, n):        csum[i] = csum[i-1] + arr[i];     # Initialize max_sm as sum of first subarray    max_sum = csum[k-1]    max_end = k-1     # Find sum of other subarrays and update    # max_sum if required.    for i in range(k, n):             curr_sum = csum[i] - csum[i-k]        if curr_sum > max_sum:                     max_sum = curr_sum            max_end = i             # Return starting index    return max_end - k + 1 # Driver programarr = [1, 12, -5, -6, 50, 3]k = 4n = len(arr)print("The maximum average subarray of length",k,"begins at index",findMaxAverage(arr, n, k)) #This code is contributed by#Smitha Dinesh Semwal

## C#

 // C# program to find maximum average// subarray of given length.using System;class GFG{ // Returns beginning index// of maximum average// subarray of length 'k'static int findMaxAverage(int []arr,                       int n, int k){         // Check if 'k' is valid    if (k > n)        return -1;     // Create and fill array    // to store cumulative    // sum. csum[i] stores    // sum of arr[0] to arr[i]    int []csum = new int[n];         csum[0] = arr[0];    for (int i = 1; i < n; i++)    csum[i] = csum[i - 1] + arr[i];     // Initialize max_sm as    // sum of first subarray    int max_sum = csum[k - 1],              max_end = k - 1;     // Find sum of other    // subarrays and update    // max_sum if required.    for (int i = k; i < n; i++)    {        int curr_sum = csum[i] -                   csum[i - k];        if (curr_sum > max_sum)        {            max_sum = curr_sum;            max_end = i;        }    }     // To avoid memory leak    //delete [] csum;         // Return starting index    return max_end - k + 1;}     // Driver Code    static public void Main ()    {        int []arr = {1, 12, -5, -6, 50, 3};        int k = 4;        int n = arr.Length;        Console.WriteLine("The maximum average subarray of "+                            "length "+ k + " begins at index "                                    + findMaxAverage(arr, n, k));    }} // This code is contributed by anuj_67.

## PHP

 \$n)        return -1;     // Create and fill array to    // store cumulative sum.    // csum[i] stores sum of    // arr[0] to arr[i]    \$csum = array();    \$csum[0] = \$arr[0];    for(\$i = 1; \$i < \$n; \$i++)    \$csum[\$i] = \$csum[\$i - 1] +                \$arr[\$i];     // Initialize max_sm as sum    // of first subarray    \$max_sum = \$csum[\$k - 1];    \$max_end = \$k - 1;     // Find sum of other subarrays    // and update max_sum if required.    for(\$i = \$k; \$i < \$n; \$i++)    {        \$curr_sum = \$csum[\$i] -                    \$csum[\$i - \$k];        if (\$curr_sum > \$max_sum)        {            \$max_sum = \$curr_sum;            \$max_end = \$i;        }    }     // Return starting index    return \$max_end - \$k + 1;}     // Driver Code    \$arr = array(1, 12, -5, -6, 50, 3);    \$k = 4;    \$n = count(\$arr);    echo "The maximum average subarray of "        ,"length ", \$k , " begins at index "        , findMaxAverage(\$arr, \$n, \$k);         // This code is contributed by anuj_67.?>

## Javascript



Output

The maximum average subarray of length 4 begins at index 1

Time Complexity: O(n), as we are using a loop to traverse n times. Where n is the number of elements in the array.
Auxiliary Space: O(n), as we are using extra space for the array csum.

We can avoid the need for extra space by using the below Efficient Method

1. Compute sum of first ‘k’ elements, i.e., elements arr[0..k-1]. Let this sum be ‘sum’. Initialize ‘max_sum’ as ‘sum’
2. Do following for every element arr[i] where i varies from ‘k’ to ‘n-1’
• Remove arr[i-k] from sum and add arr[i], i.e., do sum += arr[i] – arr[i-k]
• If new sum becomes more than max_sum so far, update max_sum.
3. Return ‘max_sum’

## C++

 // C++ program to find maximum average subarray// of given length.#includeusing namespace std; // Returns beginning index of maximum average// subarray of length 'k'int findMaxAverage(int arr[], int n, int k){    // Check if 'k' is valid    if (k > n)        return -1;     // Compute sum of first 'k' elements    int sum = arr[0];    for (int i=1; i max_sum)        {            max_sum = sum;            max_end = i;        }    }     // Return starting index    return max_end - k + 1;} // Driver programint main(){    int arr[] = {1, 12, -5, -6, 50, 3};    int k = 4;    int n = sizeof(arr)/sizeof(arr[0]);    cout << "The maximum average subarray of "         "length "<< k << " begins at index "         << findMaxAverage(arr, n, k);    return 0;}

## Java

 // Java program to find maximum average subarray// of given length. import java.io.*; class GFG {     // Returns beginning index of maximum average    // subarray of length 'k'    static int findMaxAverage(int arr[], int n, int k)    {                 // Check if 'k' is valid        if (k > n)            return -1;             // Compute sum of first 'k' elements        int sum = arr[0];        for (int i = 1; i < k; i++)            sum += arr[i];             int max_sum = sum, max_end = k-1;             // Compute sum of remaining subarrays        for (int i = k; i < n; i++)        {            sum = sum + arr[i] - arr[i-k];            if (sum > max_sum)            {                max_sum = sum;                max_end = i;            }        }             // Return starting index        return max_end - k + 1;    }     // Driver program    public static void main (String[] args)    {        int arr[] = {1, 12, -5, -6, 50, 3};        int k = 4;        int n = arr.length;        System.out.println( "The maximum average"                     + " subarray of length " + k                     + " begins at index "                    + findMaxAverage(arr, n, k));    }} // This code is contributed by anuj_67.

## Python3

 # Python 3 program to find maximum# average subarray of given length. # Returns beginning index of maximum# average subarray of length 'k'def findMaxAverage(arr, n, k):     # Check if 'k' is valid    if (k > n):        return -1     # Compute sum of first 'k' elements    sum = arr[0]         for i in range(1, k):        sum += arr[i]     max_sum = sum    max_end = k - 1     # Compute sum of remaining subarrays    for i in range(k, n):             sum = sum + arr[i] - arr[i - k]                 if (sum > max_sum):                     max_sum = sum            max_end = i             # Return starting index    return max_end - k + 1 # Driver programarr = [1, 12, -5, -6, 50, 3]k = 4n = len(arr) print("The maximum average subarray of length", k,                                "begins at index",                        findMaxAverage(arr, n, k)) # This code is contributed by# Smitha Dinesh Semwal

## C#

 // C# program to find maximum average// subarray of given length.using System; class GFG {     // Returns beginning index of    // maximum average subarray of    // length 'k'    static int findMaxAverage(int []arr,                           int n, int k)    {                 // Check if 'k' is valid        if (k > n)            return -1;             // Compute sum of first 'k'        // elements        int sum = arr[0];        for (int i = 1; i < k; i++)            sum += arr[i];             int max_sum = sum;        int max_end = k-1;             // Compute sum of remaining        // subarrays        for (int i = k; i < n; i++)        {            sum = sum + arr[i] - arr[i-k];            if (sum > max_sum)            {                max_sum = sum;                max_end = i;            }        }             // Return starting index        return max_end - k + 1;    }     // Driver program    public static void Main ()    {        int []arr = {1, 12, -5, -6, 50, 3};        int k = 4;        int n = arr.Length;        Console.WriteLine( "The maximum "          + "average subarray of length "                + k + " begins at index "            + findMaxAverage(arr, n, k));    }} // This code is contributed by anuj_67.

## PHP

 \$n)        return -1;     // Compute sum of first    // 'k' elements    \$sum = \$arr[0];    for(\$i = 1; \$i < \$k; \$i++)        \$sum += \$arr[\$i];     \$max_sum = \$sum;    \$max_end = \$k-1;     // Compute sum of    // remaining subarrays    for(\$i = \$k; \$i < \$n; \$i++)    {        \$sum = \$sum + \$arr[\$i] -                 \$arr[\$i - \$k];        if (\$sum > \$max_sum)        {            \$max_sum = \$sum;            \$max_end = \$i;        }    }     // Return starting index    return \$max_end - \$k + 1;}     // Driver Code    \$arr = array(1, 12, -5, -6, 50, 3);    \$k = 4;    \$n = count(\$arr);    echo "The maximum average subarray of ",         "length ", \$k , " begins at index "        , findMaxAverage(\$arr, \$n, \$k);         // This code is contributed by anuj_67.?>

## Javascript



Output

The maximum average subarray of length 4 begins at index 1

Time Complexity: O(n), as we are using a loop to traverse n times. Where n is the number of elements in the array.
Auxiliary Space: O(1), as we are not using any extra space.

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