Find shortest subarray with target array

Find shortest subarray with target array

/*
830pm
brute force

target [1,2,3]
arr[1,1,1,2,2,2,3,1,2,3]
                  i   j
arr [3,2,3]                  
                  
arr[1,1,1,2,2,2,3]
    i           j

X find shortest subarray that has all target numbers in same order
find shortest subarray length


*/
/*
JUlia's analysis:
  array is not sorted
  target - not duplicate, unique, ascending -> any order 
  Find shortest subarray which has all target numbers in the same order. 
  
  Brute force: 
how many subarray - start position firstIndex/ endIndex, containing 1, 2, 3,
first one, second one - 2, third one is 3 
  O(n^2) -> go through subarray once -> iterate target -> first number -> ..
  O(n)
  -> O(n^3)
  
  shortest one -> find shortest length / 
  
  Improve this algorithm: 
*/

public static int FindShortestSubArrayContainingAllKeys(int[] numbers, int[] target)
{
  if(numbers == null || target == null || numbers.Length < target.Length)
    return -1; 
  
  var nLength = numbers.Length; 
  var tLength = target.Length; 
  
  var minLength = Int.MaxValue; 
  
  for(int start = 0; start < nLength - tLength + 1; start++)  // 2 
  {    
     if(numbers[start] != target[0])
       continue; 
    
      // subarray start -> end
     // target [1,2,3]
     //arr[1,1,1,2,2,2,3,1,2,3]
     //             i   j
    //arr[1,1,1,2,2,2,3]
    //  i           j
      int tIndex = 0; 
      int tStart = 0; 
      int index = start; 
      while(index >= start && index < nLength)
      {
        var current = numbers[index]; 
        if(current == target[tIndex])
        {                    
          tIndex++; 
        }
        
        if(tIndex == 1)
          tStart = index;
        
        index++;  
        if(tIndex == tLength)
        {
          // record length - compare to minimum length
          var currentLength = index - tStart;
          minLength = currentLength < minLength? currentLength : minLength; 
        }
      }      
    }
    
    return minLength == Int.MaxValue? -1 : minLength; 
  }
}

Time complexity: (29 minutes)
should be O(n^2)
  
Brent hint: 
for(int i=0;i<arr.length; i++)
  if(arr[i]==target[0])
    for(int j=i;j<arr.length)