﻿ Java - Maximum number inside the number in the window

# Java: Find the maximum number inside the number in the window

## Java Basic: Exercise-174 with Solution

Write a Java program to find the maximum number inside the number in the window (size k) at each step in a given array of integers with duplicate numbers. Move the window to the top of the array.

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

Visual Presentation:

Sample Solution:

Java Code:

``````// Import necessary classes from java.util package
import java.util.*;
import java.util.Arrays;
// Main class to demonstrate max sliding window
public class Main {
// Main method to execute the sliding window algorithm
public static void main(String[] args) {
// Sample array and value of k for testing
int[] main_array = {1, 2, 3, 4, 5, 6, 7, 8, 8};
int k = 3;
// Display the original array and the value of k
System.out.println("\nOriginal array: " + Arrays.toString(main_array));
System.out.println("\nValue of k: " + k);
System.out.println("\nResult: ");
// Call the method to find maximums in the sliding window
ArrayList result = max_slide_window(main_array, k);
// Display the result
for (int i = 0; i < result.size(); i++) {
System.out.println(result.get(i));
}
}
// Method to find maximums in a sliding window
public static ArrayList max_slide_window(int[] main_array, int k) {
// Initialize an ArrayList to store the result
ArrayList rst_arra = new ArrayList();
// Checking for invalid inputs
if (main_array == null || main_array.length == 0 || k < 0) {
return rst_arra;
}
// Using a Deque to store indexes of elements
Deque<Integer> deque_num = new LinkedList<>();
// Processing the first k elements separately
for (int i = 0; i < k; i++) {
// Removing smaller elements from the Deque
while (!deque_num.isEmpty() && main_array[deque_num.peekLast()] <= main_array[i]) {
deque_num.pollLast();
}
deque_num.offerLast(i); // Adding the current index to the Deque
}
// Processing the rest of the elements
for (int i = k; i < main_array.length; i++) {
rst_arra.add(main_array[deque_num.peekFirst()]); // Adding the maximum from the window to result
// Removing elements that are out of the window range
if (!deque_num.isEmpty() && deque_num.peekFirst() <= i - k) {
deque_num.pollFirst();
}
// Removing smaller elements from the Deque
while (!deque_num.isEmpty() && main_array[deque_num.peekLast()] <= main_array[i]) {
deque_num.pollLast();
}
deque_num.offerLast(i); // Adding the current index to the Deque
}
rst_arra.add(main_array[deque_num.peekFirst()]); // Adding the maximum of the last window
return rst_arra; // Returning the result ArrayList containing maximums
}
}
```
```

Sample Output:

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

Value of k: 3

Result:
3
4
5
6
7
8
8
```

Flowchart:

Java Code Editor:

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