Given an array arr[] of size n and integer k such that k <= n.

Examples : 

Input:  arr[] = {3, 7, 90, 20, 10, 50, 40}, k = 3
Output: Subarray between indexes 3 and 5
The subarray {20, 10, 50} has the least average 
among all subarrays of size 3.

Input:  arr[] = {3, 7, 5, 20, -10, 0, 12}, k = 2
Output: Subarray between [4, 5] has minimum average

We strongly recommend that you click here and practice it, before moving on to the solution.

A Simple Solution is to consider every element as beginning of subarray of size k and compute sum of subarray starting with this element. Time complexity of this solution is O(nk).

An Efficient Solution is to solve the above problem in O(n) time and O(1) extra space. The idea is to use sliding window of size k. Keep track of sum of current k elements. To compute sum of current window, remove first element of previous window and add current element (last element of current window).

1) Initialize res_index = 0 // Beginning of result index
2) Find sum of first k elements. Let this sum be 'curr_sum'
3) Initialize min_sum = sum
4) Iterate from (k+1)'th to n'th element, do following
   for every element arr[i]
      a) curr_sum = curr_sum + arr[i] - arr[i-k]
      b) If curr_sum < min_sum
           res_index = (i-k+1)
5) Print res_index and res_index+k-1 as beginning and ending
   indexes of resultant subarray.

Below is the implementation of above algorithm. 

// A Simple JavaScript program to find
// minimum average subarray

// Prints beginning and ending indexes
// of subarray of size k with minimum average
function findMinAvgSubarray(arr, n, k) {

    // k must be smaller than or equal to n
    if (n < k)
        return;

    // Initialize beginning index of result
    let res_index = 0;

    // Compute sum of first subarray of size k
    let curr_sum = 0;
    for (let i = 0; i < k; i++)
        curr_sum += arr[i];

    // Initialize minimum sum as current sum
    let min_sum = curr_sum;

    // Traverse from (k+1)'th element
    // to n'th element
    for (let i = k; i < n; i++) {

        // Add current item and remove first
        // item of previous subarray
        curr_sum += arr[i] - arr[i - k];

        // Update result if needed
        if (curr_sum < min_sum) {
            min_sum = curr_sum;
            res_index = (i - k + 1);
        }
    }
    console.log("Subarray between [" + res_index +
        ", " + (res_index + k - 1) + "] has minimum average");
}

// Driver code
let arr = [3, 7, 90, 20, 10, 50, 40];

// Subarray size
let k = 3;
let n = arr.length;

findMinAvgSubarray(arr, n, k);

// A Simple JavaScript program to find

// minimum average subarray

​

// Prints beginning and ending indexes

// of subarray of size k with minimum average

function findMinAvgSubarray(arr, n, k) {

​

    // k must be smaller than or equal to n

    if (n < k)

        return;

​

    // Initialize beginning index of result

    let res_index = 0;

​

    // Compute sum of first subarray of size k

    let curr_sum = 0;

    for (let i = 0; i < k; i++)

        curr_sum += arr[i];

​

    // Initialize minimum sum as current sum

    let min_sum = curr_sum;

​

    // Traverse from (k+1)'th element

    // to n'th element

    for (let i = k; i < n; i++) {

​

        // Add current item and remove first

        // item of previous subarray

        curr_sum += arr[i] - arr[i - k];

​

        // Update result if needed

        if (curr_sum < min_sum) {

            min_sum = curr_sum;

            res_index = (i - k + 1);

        }

    }

    console.log("Subarray between [" + res_index +

        ", " + (res_index + k - 1) + "] has minimum average");

}

​

// Driver code

let arr = [3, 7, 90, 20, 10, 50, 40];

​

// Subarray size

let k = 3;

let n = arr.length;

​

findMinAvgSubarray(arr, n, k);

Output: 

Subarray between [3, 5] has minimum average

Complexity Analysis:

• Time Complexity: O(n) 
• Auxiliary Space: O(1)

Please refer complete article on Find the subarray with least average for more details!

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Article Tags :

• Arrays
• DSA
• JavaScript
• Mathematical
• Web Technologies
• Amazon

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