Array distinct element - find value equal the index of element using binary search, need to remove diff extra array, so time complexity can lower to O(logn) instead of O(n).

ArrayDistinct_FindValueEqualIndex.cs

using System;

class Solution
{
    public static int IndexEqualsValueSearch(int[] arr) // [0, 1, 2, 3]
    {
      if(arr == null || arr.Length == 0) // false 
      {
        return -1; 
      }
      
      var length = arr.Length; // 4
      var diff = new int[length]; // O(n)
      for(int i = 0; i < length; i++)  // bottom - O(n)
      {
        diff[i] = arr[i] - i; // 0
      }
      
      return binarySearch(diff,0,  0, length - 1); // 0, 0 , 3
    }
  
    private static int binarySearch(int[] numbers, int search, int start, int end)  // logn - 
    {
      if(start > end)  // false 
      {
        return -1; 
      }
      
      var middle = start + (end - start)/2; // 1
      var middleValue = numbers[middle];  // 0 
      // base case 
      if(middleValue == search) // 
      {
        // You need to make some changes here
        // You definitely know 1 index
        // [0 0 0 0 0]
        // middle = 2
        // [0 0]
        if(middle > 0 && numbers[middle - 1] == 0)
        {
          return binarySearch(numbers, search, start, middle - 1);         
        }
        else 
        {
          return middle; // 1 -> lowest one - bug fix 
        }
      }
      else if( middleValue < search)
      {
        start = middle + 1; 
      }
      else 
      {
        end = middle - 1; 
      }
      
      return binarySearch(numbers, search, start, end); 
    }

    static void Main(string[] args)
    {

    }
}

// [-8, 0, 2, 5] - distinct ascending -> increment at least 1
// [0, 1, 2, 3]  -> ascending order -> increment 1 
// arr[i] - i -> array not descending 
// arr[i] <= arr[i] if i < j
// [0, 1, 2, 3]
// newArray = arr[i] - i , saying that this array is sorted, 
// [0,0,0,0] -> sorted, 
// argument if it is sorted, binary search 
// O(logn)  given value 0, find index - arr[i] - i = 0, 
//  space O(1)