Find the duplicate - two pointer technique

FindTheDuplicate_TwoPointerTechnique.cs

using System;
using System.Collections.Generic;

class Solution
{
    public static int[] FindDuplicates(int[] arr1, int[] arr2) // [1, 2, 3,5, 6,7], [3, 6,7,8, 10]
    {
      if(arr1 == null || arr1.Length == 0 || arr2 == null || arr2.Length == 0)  // false 
      {
        return new int[0];
      }
      
      int length1 = arr1.Length; // 6
      int length2 = arr2.Length; // 5
      
      int start1 = 0; 
      int start2 = 0; 
      
      var list = new List<int>(); 
      
      while(start1 < length1 && start2 < length2) // true
      {
        var visit1 = arr1[start1]; // 1, 2, 3
        var visit2 = arr2[start2]; // 3
        
        var isSame = visit1 == visit2; // false 
        if(isSame) // true
        {
          list.Add(visit1); // 3, 
          start1++; // 3
          start2++; // 2
        }
        else if(visit1 < visit2)
        {
          start1++; // 1, 2
        }
        else 
        {
          start2++; 
        }
      }
      
      // LINQ statement  - C# 
      // https://stackoverflow.com/questions/629178/conversion-from-listt-to-array-t
      /*
      int length = list.Count; 
      
      var result = new int[length]; 
      for(int i = 0; i < length; i++)
      {
        result[i] = list[i]; 
      }
      
      return result; 
      */
      return list.ToArray(); // String.ToArray()  ASP.NET abstract class, interface, API - IEnemerable, ICollection
    }

    static void Main(string[] args)
    {
         var duplicate = FindDuplicates(new int[]{1, 2,3,5,6, 7}, new int[]{3, 6, 7, 8, 20}); 
        foreach(var item in duplicate)
        {
           Console.WriteLine(item); 
        }
    }
}

// linear scan 
// O(m + n) - two array sizes are almost the same , first case
// [1, 2, 3, 5, 6, 7]  -> [3, ]
/*
             |
    
   [3, 6, 7, 8, 20]
       |        
*/
/*
M >> N

work on N, small array [1, 2, 3]
1, try to find it in the second array using binary search log(N)
My solution using binary search is N * log(M), it is better than brute force M * N
*/