Median of sliding window in an array

13 Jan, 2020
Given an array of integer arr[] and an integer k, the task is to find the median of each window of size k starting from the left and moving towards the right by one position each time.
Examples:
Input: arr[] = {-1, 5, 13, 8, 2, 3, 3, 1}, k = 3Output: 5 8 8 3 3 3
Input: arr[] = {-1, 5, 13, 8, 2, 3, 3, 1}, k = 4Output: 6.5 6.5 5.5 3.0 2.5
Approach: Create a pair class to hold the items and their index. It also implements the comparable interface so that compareTo() method will be invoked by the Treeset to find the nodes. Note that the two pairs are equal only when their indices are equal. This is important since a window can contain duplicates and we may end up deleting multiple items in single remove() call if we only check for the value.
The idea is to maintain two sorted sets (minSet and maxSet) of Pair objects of length (k / 2) and (k / 2) + 1 depending on whether k is even or odd, minSet will always contain the first set of numbers (smaller) of window k and maxSet will contain the second set of numbers (larger).
As we move our window, we will remove elements from either of the sets (log n) and add a new element (log n) maintaining the minSet and maxSet rule specified above.
Below is the implementation of the above approach:
// Java implementation of the approachimport java.util.TreeSet;  public class GFG {      // Pair class for the value and its index    static class Pair implements Comparable<Pair> {        private int value, index;          // Constructor        public Pair(int v, int p)        {            value = v;            index = p;        }          // This method will be used by the treeset to        // search a value by index and setting the tree        // nodes (left or right)        @Override        public int compareTo(Pair o)        {              // Two nodes are equal only when            // their indices are same            if (index == o.index) {                return 0;            }            else if (value == o.value) {                return Integer.compare(index, o.index);            }            else {                return Integer.compare(value, o.value);            }        }          // Function to return the value        // of the current object        public int value()        {            return value;        }          // Update the value and the position        // for the same object to save space        public void renew(int v, int p)        {            value = v;            index = p;        }          @Override        public String toString()        {            return String.format("(%d, %d)", value, index);        }    }      // Function to print the median for the current window    static void printMedian(TreeSet<Pair> minSet,                            TreeSet<Pair> maxSet, int window)    {          // If the window size is even then the        // median will be the average of the        // two middle elements        if (window % 2 == 0) {            System.out.print((minSet.last().value()                              + maxSet.first().value())                             / 2.0);            System.out.print(" ");        }          // Else it will be the middle element        else {            System.out.print(minSet.size() > maxSet.size()                                 ? minSet.last().value()                                 : maxSet.first().value());            System.out.print(" ");        }    }      // Function to find the median    // of every window of size k    static void findMedian(int arr[], int k)    {        TreeSet<Pair> minSet = new TreeSet<>();        TreeSet<Pair> maxSet = new TreeSet<>();          // To hold the pairs, we will keep renewing        // these instead of creating the new pairs        Pair[] windowPairs = new Pair[k];          for (int i = 0; i < k; i++) {            windowPairs[i] = new Pair(arr[i], i);        }          // Add k/2 items to maxSet        for (int i = 0; i < k / 2; i++) {            maxSet.add(windowPairs[i]);        }          for (int i = k / 2; i < k; i++) {              // Below logic is to maintain the            // maxSet and the minSet criteria            if (arr[i] < maxSet.first().value()) {                minSet.add(windowPairs[i]);            }            else {                minSet.add(maxSet.pollFirst());                maxSet.add(windowPairs[i]);            }        }          printMedian(minSet, maxSet, k);          for (int i = k; i < arr.length; i++) {              // Get the pair at the start of the window, this            // will reset to 0 at every k, 2k, 3k, ...            Pair temp = windowPairs[i % k];            if (temp.value() <= minSet.last().value()) {                  // Remove the starting pair of the window                minSet.remove(temp);                  // Renew window start to new window end                temp.renew(arr[i], i);                  // Below logic is to maintain the                // maxSet and the minSet criteria                if (temp.value() < maxSet.first().value()) {                    minSet.add(temp);                }                else {                    minSet.add(maxSet.pollFirst());                    maxSet.add(temp);                }            }            else {                maxSet.remove(temp);                temp.renew(arr[i], i);                  // Below logic is to maintain the                // maxSet and the minSet criteria                if (temp.value() > minSet.last().value()) {                    maxSet.add(temp);                }                else {                    maxSet.add(minSet.pollLast());                    minSet.add(temp);                }            }              printMedian(minSet, maxSet, k);        }    }      // Driver code    public static void main(String[] args)    {        int[] arr = new int[] { 0, 9, 1, 8, 2,                                7, 3, 6, 4, 5 };        int k = 3;          findMedian(arr, k);    }}
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median-finding
sliding-window
Arrays
Binary Search Tree
Dynamic Programming
Java Programs
Sorting
sliding-window
Arrays
Dynamic Programming
Sorting
Binary Search Tree
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