# Find maximum average subarray of k length

Given an array with positive and negative numbers, find the maximum average subarray of 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
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A Simple Solution is to run two loops. The outer loop picks starting point, the inner loop goes till length ‘k’ from the starting point and computes average of elements. Time complexity of this solution is O(n*k).

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 csum[] array with us, we can compute sum between two indexes in O(1) time.
Below is C++ implementation of this idea. One observation is, a subarray of given length has maximum average if it has maximum sum. So we can avoid floating point arithmetic by just comparing sum.

```// C++ program to find maximum average subarray
// of given length.
#include<bits/stdc++.h>
using 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<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;
}
}

delete [] csum; // To avoid memory leak

// Return starting index
return max_end - k + 1;
}

// Driver program
int 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;
}
```

Output:

`The maximum average subarray of length 4 begins at index 1`

Time Complexity of above solution is O(n), but it requires O(n) auxiliary space.

We can avoid need of extra space by using 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’
…….a) Remove arr[i-k] from sum and add arr[i], i.e., do sum += arr[i] – arr[i-k]
…….b) If new sum becomes more than max_sum so far, update max_sum.
3) Return ‘max_sum’

```// C++ program to find maximum average subarray
// of given length.
#include<bits/stdc++.h>
using 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<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++)
{
int 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
int 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;
}```

Output:

`The maximum average subarray of length 4 begins at index 1`

Time complexity of this method is also O(n), but it requires constant extra space.

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