Sir my code is correct but it is too big so can u make this shorter

#1

import java.util.Scanner;

class Patterntry {
static Scanner scn = new Scanner(System.in);

public static void main(String args[]) {
	int n = scn.nextInt();
	int[] arr1 = new int[n];
	insert(arr1);
	int[] arr2 = new int[n];
	insert(arr2);
	int[] arr = new int[2 * n];
	for (int j = 0; j < arr.length;) {
		for (int i = 0; i < arr1.length; i++) {
			arr[j]=arr1[i];
			j++;
		}
		for (int i = 0; i < arr2.length; i++) {
			arr[j]=arr2[i];
			j++;
		}
	}bublesort(arr);
	System.out.println(arr[(arr.length-1)/2]);
}

public static void insert(int[] arr) {
	for (int i = 0; i < arr.length; i++) {
		arr[i] = scn.nextInt();
	}
}

public static void bublesort(int[] arr) {
	for (int c = 0; c < arr.length - 1; c++) {
		for (int i = 0; i < arr.length - 1 - c; i++) {
			if (arr[i] > arr[i + 1]) {
				int temp = arr[i];
				arr[i] = arr[i + 1];
				arr[i + 1] = temp;
			}
		}
	}
}

}

#2

@Harsh_Sonwani,
You are currently using a brute force approach which has a high time complexity but its correct.

Here is an alternate approach:

  1. Calculate the medians m1 and m2 of the input arrays ar1[] and ar2[] respectively.

  2. If m1 and m2 both are equal then we are done. return m1 (or m2) .

  3. If m1 is greater than m2, then median is present in one of the below two subarrays.
    a) From first element of ar1 to m1 (ar1[0…| n/2 |]) .
    b) From m2 to last element of ar2 (ar2[| n/2 |…n-1]) .

  4. If m2 is greater than m1, then median is present in one
    of the below two subarrays.
    a) From m1 to last element of ar1 (ar1[| n/2 |…n-1]) .
    b) From first element of ar2 to m2 (ar2[0…| n/2 |])

  5. Repeat the above process until size of both the subarrays becomes 2.

  6. If size of the two arrays is 2 then use below formula to get the median.
    Median = (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1]))/2

#3

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