Java Exercises: Find the median of the number inside the window

Java Basic: Exercise-173 with Solution

Write a Java program to find the median of the number inside the window (size k) at each moving in a given array of integers with duplicate numbers. Move the window from the start of the array.

{|1, 2, 3|, 4, 5, 6, 7, 8, 8} -> Return median 2
{1, |2, 3, 4|, 5, 6, 7, 8, 8} -> Return median 3
{1, 2, |3, 4, 5|, 6, 7, 8, 8} -> Return median 4
{1, 2, 3, |4, 5, 6|, 7, 8, 8} -> Return median 5
{1, 2, 3, 4, |5, 6, 7|, 8, 8} -> Return median 6
{1, 2, 3, 4, 5, |6, 7, 8|, 8} -> Return median 7
{1, 2, 3, 4, 5, 6, |7, 8, 8|} -> Return median 8
Result array {2, 3, 4, 5, 6, 7, 8}

Sample Solution:

Java Code:

import java.util.*;
import java.util.Arrays;
import java.util.LinkedList;

public class Solution {
 public static void main(String[] args) {
  int[] main_array = {1, 2, 3, 4, 5, 6, 7, 8, 8};
  int k = 3;
  System.out.println("\nOriginal array: " + Arrays.toString(main_array));
  System.out.println("\nValue of k: " + k);
  System.out.println("\nResult: ");
  ArrayList < Integer > result = median_slide_window(main_array, k);
  for (int i = 0; i < result.size(); i++) {
   System.out.println(result.get(i));
  }
 }
 public static ArrayList < Integer > median_slide_window(int[] main_array, int k) {
  ArrayList < Integer > result = new ArrayList < > ();
  if (k == 0 || main_array.length < k) {
   return result;
  }
  PriorityQueue < Integer > right_num = new PriorityQueue < Integer > (k);
  PriorityQueue < Integer > left_num = new PriorityQueue < Integer > (k, Collections.reverseOrder());

  for (int i = 0; i < k - 1; ++i) {
   add(right_num, left_num, main_array[i]);
  }

  for (int i = k - 1; i < main_array.length; ++i) {
   add(right_num, left_num, main_array[i]);
   int median = compute_median(right_num, left_num);
   result.add(median);
   remove(right_num, left_num, main_array[i - k + 1]);
  }
  return result;
 }

 private static int compute_median(PriorityQueue < Integer > right_num, PriorityQueue < Integer > left_num) {
  if (left_num.isEmpty() && right_num.isEmpty()) {
   return 0;
  }
  while (left_num.size() < right_num.size()) {
   left_num.add(right_num.poll());
  }

  while (left_num.size() - right_num.size() > 1) {
   right_num.add(left_num.poll());
  }
  return left_num.peek();
 }

 private static void add(PriorityQueue < Integer > right_num, PriorityQueue < Integer > left_num, int num) {
  if (left_num.isEmpty() && right_num.isEmpty()) {
   left_num.add(num);
   return;
  } else {
   if (num <= compute_median(right_num, left_num)) {
    left_num.add(num);
   } else {
    right_num.add(num);
   }
  }
 }

 private static void remove(PriorityQueue < Integer > right_num, PriorityQueue < Integer > left_num, int num) {
  if (num <= compute_median(right_num, left_num)) {
   left_num.remove(num);
  } else {
   right_num.remove(num);
  }
 }
}

Sample Output:

Original array: [1, 2, 3, 4, 5, 6, 7, 8, 8]

Value of k: 3

Result: 
2
3
4
5
6
7
8

Flowchart:

Java Code Editor:

Company:  Google

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