Question

Given an array ? of numbers and a window size ?, the sliding window ending at index ? is the subarray ?[? − ? + 1], ⋯ , ?[?] if ? ≥ ? − 1, and ?[0], ⋯ , ?[?] otherwise. For example, if ? = [0, 2, 4, 6, 8] and ? = 3, then the sliding windows ending at indices 0, 1, 2, 3 and 4 are respectively [0], [0, 2], [0, 2, 4], [2, 4, 6], and [4, 6, 8]. Write a method, movingAverage, that given as input an array ? of numbers and a window size ?, returns an array ? with the same length as ?, such that for every index ?, ?[?] is the average of the numbers in the sliding window ending at index ?. The skeleton for the method is provided in the file MovingAverage.java.

The following is a sample run.

Input:      ? = [0,2, 4, 6, 8],   ? = 3

Return:   [0, 1, 2, 4, 6]

Explanation: As explained above, the sliding windows at indices 0, 1, 2, 3 and 4 are respectively

[0], [0,2], [0, 2, 4], [2, 4, 6], and [4, 6, 8]. The average of the numbers in these sliding windows are respectively = 0, = 1, = 2, = 4, and = 6.

Your method must have time complexity ?(?), where ? is the length of the input array ?.

Hint: You may find a queue useful.

public class MovingAverage {

   public double[] movingAverage(int[] A, int w) {
      
       //Replace this line with your return statement
       return null;
   }

}

Answer 1

        public static double[] movingAverage(int[] A, int w) {
                
                double result[] = new double[A.length];
                
                double currentSum = 0;
                for(int i=0; i<A.length; i++) {
                        currentSum += A[i];
                        
                        if(i >= w) {
                                currentSum -= A[i-w];
                        }
                        
                        int len = Math.min(i + 1, w);
                        result[i] = currentSum/len;
                }

                // Replace this line with your return statement
                return result;
        }

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