Find The Duplicates

[esthicodes]

Given two sorted arrays arr1 and arr2 of passport numbers, implement a function findDuplicates that returns an array of all passport numbers that are both in arr1 and arr2. Note that the output array should be sorted in an ascending order.

Let N and M be the lengths of arr1 and arr2, respectively. Solve for two cases and analyze the time & space complexities of your solutions: M ≈ N - the array lengths are approximately the same M ≫ N - arr2 is much bigger than arr1.

Example:

input: arr1 = [1, 2, 3, 5, 6, 7], arr2 = [3, 6, 7, 8, 20]

output: [3, 6, 7] # since only these three values are both in arr1 and arr2
Constraints:

[time limit] 5000ms

[input] array.integer arr1

1 ≤ arr1.length ≤ 100
[input] array.integer arr2

1 ≤ arr2.length ≤ 100
[output] array.integer

[esthicodes]

Wrong Example Code:

import java.io.*;
import java.util.*;

class Solution {

  static int[] binarySearch(int[] arr, int num) {
    // your code goes here
    int begin = 0;
    int end = arr.length -1;
    
    while(begin <= end) {
      int mid = begin + (end - begin / 2);
      if (arr[mid] < num) 
        begin = mid + 1;
      else if (arr[mid] == num)
        return mid;
      else 
        end = mid - 1 ;
      
    }
    
    return -1;
  }
  
  static int[] findDuplicates(int[] arr1, int[] arr2) {
    int[] duplicates = new int [arr1.length];
    
    int j = 0;
    for (int i = 0 ; i < arr1.length; i++) {
      if(binarySearch(arr2, arr1[i]) != -1)
        
        return Arrays.copyOf(duplicates, j);
    }
    public static void main(String[] args) {
      
    }
  }
      duplicates[j++] = arr1[i];
}