Items - Finding

If your array contains less than 100,000 elements then looping through once is very efficient.
Exactly which method you use will depend on the following questions:
1) Is the array 1 dimensional or multi-dimensional.
2) Are you looking to conduct an exact match or just a partial match.
3) Do you need to know the position of the item if it exists.
4) How many searches do you need to do, it might be worth sorting the array.
5) How often is the data changing, you may need to keep re-sorting.

Exact Match

If you are using Excel VBA you can use the Excel MATCH worksheet function.

Public Function OneDimensional_Exact_One( _ 
         ByVal arValues As Variant, _ 
         ByVal sFindValue As String) _ 
         As Boolean 

   On Error GoTo ErrorHandler 
   OneDimensional_Exact_One = Not IsError(Application.WorksheetFunction.Match(sFindValue, arValues, 0)) 
   Exit Function 

ErrorHandler: 
   OneDimensional_Exact_One = False 
End Function 

You can use the FILTER function to tell if an item exists in a one-dimensional array.
This function returns an array of any elements that contain a given text string.
This function takes a string array, text string and returns a one-dimensional array containing all the elements that match the search string.
This method does not distinguish between complete matches and substring matches.
This method does not work with numerical arrays since all the numbers have to be implicitly converted to text first

Public Function OneDimensional_Exact_Two( _ 
         ByVal arValues As Variant, _ 
         ByVal sFindValue As String) _ 
         As Boolean 
   
   OneDimensional_Exact_Two = (UBound(VBA.Filter(arValues, sFindValue)) > -1) 
End Function 

You can use the INSTR function after converting the array to a string concatenation
This method also uses the JOIN function to return a text string containing all the elements in an array (String).

Public Function OneDimensional_Exact_Three( _ 
         ByVal arValues As Variant, _ 
         ByVal sFindValue As String) _ 
         As Boolean 

Dim stextconcat As String 
Dim sdelimiter As String 

   sdelimiter = "--" 
   stextconcat = sdelimiter & VBA.Join(arValues, sdelimiter) & sdelimiter 
   OneDimensional_Exact_Three = VBA.InStr(1, stextconcat, sdelimiter & sFindValue & sdelimiter) 
   
   'use this line for case insensitive searches
   'OneDimensional_Exact_Three = VBA.InStr(1, stextconcat, sdelimiter & sFindValue & sdelimiter, vbTextCompare)
End Function 

You can loop through the array using a For Next - Each loop.

Public Function OneDimensional_Exact_Four( _ 
         ByVal arValues As Variant, _ 
         ByVal sFindValue As String) _ 
         As Boolean 

Dim sItem As Variant 
   
   For Each sItem In arValues 
      If sItem = sFindValue Then 
          OneDimensional_Exact_Four = True 
          Exit Function 
      End If 
   Next sItem 
End Function 

Partial Match

This function uses a 1-dimensional array starting at 1.

Public Sub Testing() 
Dim myArray(1 To 4) As String 
Dim bfoundMatch As String 

   myArray(1) = "one1" 
   myArray(2) = "two3" 
   myArray(3) = "three5" 
   myArray(4) = "four7" 
   
   bfoundMatch = OneDimensional_Partial(myArray, "two") 
   Debug.Print bfoundMatch 
End Sub 

Public Function OneDimensional_Partial(ByVal arValues As Variant, _ 
                                       ByVal sFindValue As String) _ 
                                       As Boolean 
Dim lrow As Long 
Dim lcolumn As Long 
    
   For lrow = LBound(arValues, 1) To UBound(arValues, 1) 
      
      If (VBA.InStr(arValues(lrow), sFindValue) > 0) Then 
         OneDimensional_Partial = True 
         Exit Function 
      End If 
         
   Next lrow 
End Function 

Multi Dimensional - Partial Match

This function uses a 2-dimensional array starting at 1.

Public Sub Testing() 
Dim myArray(1 To 4, 1 To 2) As String 
Dim bfoundMatch As String 

    myArray(1, 1) = "one1" 
    myArray(1, 2) = "one2" 
    myArray(2, 1) = "two3" 
    myArray(2, 2) = "two4" 
    myArray(3, 1) = "three5" 
    myArray(3, 2) = "three6" 
    myArray(4, 1) = "four7" 
    myArray(4, 2) = "four8" 
    
    bfoundMatch = TwoDimensional_Partial(myArray, "two") 
    Debug.Print bfoundMatch 
End Sub 

Public Function TwoDimensional_Partial( _ 
         ByVal arValues As Variant, _ 
         As Boolean 

   Dim lrow As Long 
   Dim lcolumn As Long 
    
   For lrow = LBound(arValues, 1) To UBound(arValues, 1) 
      For lcolumn = LBound(arValues, 2) To UBound(arValues, 2) 
      
         If (VBA.InStr(arValues(lrow, lcolumn), sFindValue) > 0) Then 
            TwoDimensional_Partial = True 
            Exit Function 
         End If 
         
      Next lcolumn 
   Next lrow 
End Function 

Binary Search

Also known as half-interval search, logarithmic search, or binary chop.
A binary search is a search algorithm that can find the position of a value within a sorted array.
The function does not return the position, it just returns whether a match was found or not.
Your array must be sorted in Ascending Order for this function to work.

Public Sub BinarySearch() 
    
Dim myArray(0 To 3) As String 
Dim bfoundMatch As String 

    myArray(0) = "four" 
    myArray(1) = "one" 
    myArray(2) = "three" 
    myArray(3) = "two" 

    bfoundMatch = Array_Find(myArray, "three") 
    Debug.Print bfoundMatch 
End Sub 

Public Function Array_Find(ByRef theArray() As String, _ 
                           ByVal target As String) _ 
                           As Boolean 

Dim bfind As Boolean 
Dim low As Integer 
Dim high As Integer 
Dim middle As Integer 
    
    low = 0 
    high = UBound(theArray) 

    bfind = False 
    Do While low <= high 
        middle = (low + high) / 2 

        If target = theArray(middle) Then 
            bfind = True 
            Exit Do 

        ElseIf target < theArray(middle) Then 
            high = (middle - 1) 
        Else 
            low = (middle + 1) 
        End If 
    Loop 
    
    Array_Find = bfind 
End Function