Maximum average subarray of size k

Given an integer array and a number k, print the maximum sum subarray of size k.
Maximum average subarray of size k is a subarray (sequence of contiguous elements in the array) for which the average is maximum among all subarrays of size k in the array.
Average of k elements = (sum of k elements)/k
Since k > 0, the maximum sum subarray of size k will also be maximum average subarray of size k. So, the problem reduces to finding maximum sum subarray of size k in the array.

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We use sliding window strategy for solving this.
Step 1: Find sum of first k elements in the input array. Initialize maxSum to the calculated sum and maxSumStartIndex = 0.
Step 2: Add next element to the sum and subtract first element from the sum. Check if this sum is greater than previous sum and update maxSum and maxSumStartIndex.
Step 3: Keep adding next element to the sum and removing first element from the sum to get sum of current sub array of size k and update maxSum and maxSumStartIndex whenever a greater sum is seen.
Step 4: Finally print k elements starting from maxSumStartIndex.

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Algorithm Visualization

Code Snippet

package com.ideserve.questions.saurabh;

 * <b>IDeserve <br>
 * <a href=""></a>
 * Given an integer array and a number k, print the maximum sum subarray of size k.
 * @author Saurabh
public class MaxAvgSubarray {

    private static int getMaxAvgSubarrayStartIndex(int input[], int k)
        int n = input.length;
        if (k > n)
            throw new IllegalArgumentException("k should be less than or equal to n");
        if(k == n) {
            return 0;   // whole array is the solution

        // Sum of first k elements
        int sum = input[0];
        for (int i = 1; i < k; i++)
            sum += input[i];
        // Initialized to first k elements sum
        int maxSum = sum;
        int maxSumIndex = 0;
        // Sum of remaining sub arrays
        for (int i = k; i < n; i++)
            // Remove first element of the current window and add next element to the window
            sum = sum - input[i-k] + input[i] ;
            if (sum > maxSum)
                maxSum = sum;
                maxSumIndex = i-k;
        // Return starting index of max avg sub array
        return maxSumIndex + 1;
    public static void printMaxAvgSubarray(int[] input, int k) {
        System.out.print("Maximum average subarray of length " + k + " is:  ");
        int index = getMaxAvgSubarrayStartIndex(input, k);
        for(int i =0 ; i < k; i++) {
            System.out.print(input[index++] + " ");
    public static void main(String[] args) {
        int[] input = {11, -8, 16, -7, 24, -2, 3};
        int k = 3;
        printMaxAvgSubarray(input, k);

Order of the Algorithm

Time Complexity is O(n)
Space Complexity is O(1)


  • Sincere thanks from IDeserve community to Saurabh Kumar for compiling current post.

    Saurabh Kumar

    Ninja Programmer