[Java] Excercise: compare two given arrays and determine their equality

Motivation: In Java, check if Two Arrays are Equal or not Equal.

a={1,2,3,4,5,6}

b={2,3,1,4,5,4,4}

Time complexity: $O(n^2)$ (because of the bubble sort algorithm)

- Version 1: 1 process. double-nested loop statement.

- Version 2: 2 processes. 1) bubble sort algorithm: $O(n^2)$, 2) single-nested loop statement O(n)

- Version 3: 1 process. 1) Arrays.equals(a, b): $O(n^2)$ (still $n^2$)

- Version 4: 2 processes. 1) check lengths of arrays, 2) sort arrays, 3) compare elements of arrays. Still $O(n^2)$ because of sorting arrays.

Java codes

import java.util.Arrays;

public static void main(String[] args) {
		
	int a[] = {1,2,3,4,5,6};
	int b[] = {2,3,1,4,5,4,4};
	boolean result = true;
	

	// version 1
	// determine whether two arrays are equal, by comparing elements one-by-one: version 1 
	// this is theta(n^2) algorithm
	for (int i = 0; i < a.length; i++) {			
		for (int j = 0; j < b.length; j++) {				
			if (a[i]!=b[j]) {
				result = false;
			}				
		}			
	}		
	// Display result (true or false)
	System.out.print("Version 1: result using double nested loop statements (O(n^2)): " );
	System.out.println(result);
		
	// sort arrays: version 2
	// this is theta(n^2) algorithm
	// sort array a
	for (int i = 0; i < a.length; i++) {

		for (int j = 0; j < a.length - 1; j++) {

			if (a[j] > a[j+1]) {

				// swap
				int temp = a[j];
				a[j] = a[j+1];
				a[j+1] = temp;					
			}				
		}			
	}

	// sort array b 
	for (int i = 0; i < b.length; i++) {
			
		for (int j = 0; j < b.length - 1; j++) {
				
			if (b[j] > b[j+1]) {
					
				// swap
				int temp = b[j];
				b[j] = b[j+1];
				b[j+1] = temp;					
			}				
		}			
	}
		
	// print array a
	System.out.print("Sorted array a: ");
	for (int i = 0; i < a.length; i++) {
		System.out.print(a[i]);			
	}		
		
	// make a new line
	System.out.println();
		
	// print array b		
	System.out.print("Sorted array b: ");
	for (int i = 0; i < b.length; i++) {
		System.out.print(b[i]);			
	}		
		
		
	// determine whether two arrays are equal, by comparing elements one-by-one: version 2
	// this is theta(n) algorithm, given that arrays are sorted
	int i = 0;
	int j = 0;		
		
	while (i < a.length && j < b.length) {
			
		// elements are equal
		if (a[i] == b[j]) {
			i = i + 1;
			j = j + 1;
		}
		else {
			result = false;
			break;
		}
		result = true;
	}
		
		System.out.println();
		System.out.println("Version 2: result using 1) sorted Arrays (O(n^2)), 2) a single nested  loop (O(n)). 3) still final time complexity is O(n^2) : "); 
		System.out.print(result); // false if output is 1, true if 0.  
		System.out.println();
		
	// Version 3: 
	boolean result2 = Arrays.equals(a, b);
        System.out.println("Result using Arrays.equals(a,b): " + result2);
        
        // Version 4: 
        boolean result3 = true;
    	int c[] = {2,3,1,4,5,4,4};

        if (a.length != c.length) {
            result = false;
        } else {
            // Sort arrays
            Arrays.sort(a);
            Arrays.sort(c);

            // Check if arrays are equal
            for (int index = 0; index < a.length; index++) {
                if (a[index] != c[index]) {
                    result = false;
                    break;
                }
            }
        }

        System.out.println("Result using Arrays.sort: " + result);      		
} // end of main