Next letter in sequence

Hi All,

I have a report in Excel 2010 which summarises costs against certain activity charge codes. The charge codes for this particular project are sequential and always 8 alpha characters. They all start with the same 4 characters and then end with AAAA, moving on to AAAB, AAAC etc. I also use this list to indicate which activity charge code is needed next in the sequence and so far this works.

The summary list has a column that pulls the last character for each of the activities, populating a range G2:G26. I then use the following formula to determine what character I need next:

=CHAR(MAX(CODE(G2:G26))+1)  entered as an array formula.

My list of last characters is not in alphabetical order so it isn't as simple as looking at the last one in the list. The current list has reached Y so the next activity to be raised will end in AAAZ. Thats fine, my formula currently returns Z as the correct answer.

However when I go beyond Z, the next activity will have to end AABA, then AABB etc.

A sample of the set up is attached. You will see from the sample that some tasks have multiple phases (preparation and execution (P / E)) and each phase has a new activity code, hence why they are not in alphabetical order. Some tasks were raised in original group and then others have been added.

The formula that I am trying to update is highlighted. I am happy to change the formulas to pull out the existing suffixes if required, hence why I have supplied the full suite of data as best I can.

At the moment I can do the increment simply but as this project expands it will become less simple.

Thanks
Rob H
Sequential-lettering.xlsx
LVL 36
Rob HensonFinance AnalystAsked:
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Michael FowlerSolutions ConsultantCommented:
I couldn't think of a way to do this with a formula, while that does not mean it cannot be done, here is a VBA function to do it for you

Function LetterIncrement(stringCounter As String) As String
    Dim arr(1 To 4) As Integer
    Dim i As Integer
    
    For i = 1 To 4
        arr(i) = Asc(UCase(Mid(stringCounter, i, 1)))
    Next
    
    arr(4) = arr(4) + 1
    
    For i = 4 To 1 Step -1
        If (arr(i) > 90) Then
            arr(i) = 65
            arr(i - 1) = arr(i - 1) + 1
        End If
    Next
    
    LetterIncrement = Chr(arr(1)) & Chr(arr(2)) & Chr(arr(3)) & Chr(arr(4))
    
End Function

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Note:
I deliberately let it throw an error if "ZZZZ" is entered
Always returns upper case but will take upper or lower as an argument
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Rob HensonFinance AnalystAuthor Commented:
Thanks I will check it out when back at desk tomorrow.

Question though, this looks like it needs an input of 4 letters in a string and it will give the next string of 4. How do I determine the string of 4 to enter? I can change my list so that it shows 4 characters rather than 1 but how can I check the list for the "highest" and thus get the next.

Thanks
0
Ejgil HedegaardCommented:
It is possible to make an array formula to do it.
I based it on the Charge code column D on sheet Data, looking at the range G2:G1000.

=CHAR(MOD(INT((MAX((IFERROR(CODE(MID($G2:$G$1000,5,1))-65,0))*26^3+(IFERROR(CODE(MID($G2:$G$1000,6,1))-65,0))*26^2+(IFERROR(CODE(MID($G2:$G$1000,7,1))-65,0))*26+IFERROR(CODE(MID($G2:$G$1000,8,1))-65,0))+1)/26^3),26)+65)&CHAR(MOD(INT((MAX((IFERROR(CODE(MID($G2:$G$1000,5,1))-65,0))*26^3+(IFERROR(CODE(MID($G2:$G$1000,6,1))-65,0))*26^2+(IFERROR(CODE(MID($G2:$G$1000,7,1))-65,0))*26+IFERROR(CODE(MID($G2:$G$1000,8,1))-65,0))+1)/26^2),26)+65)&CHAR(MOD(INT((MAX((IFERROR(CODE(MID($G2:$G$1000,5,1))-65,0))*26^3+(IFERROR(CODE(MID($G2:$G$1000,6,1))-65,0))*26^2+(IFERROR(CODE(MID($G2:$G$1000,7,1))-65,0))*26+IFERROR(CODE(MID($G2:$G$1000,8,1))-65,0))+1)/26),26)+65)&CHAR(MOD((MAX((IFERROR(CODE(MID($G2:$G$1000,5,1))-65,0))*26^3+(IFERROR(CODE(MID($G2:$G$1000,6,1))-65,0))*26^2+(IFERROR(CODE(MID($G2:$G$1000,7,1))-65,0))*26+IFERROR(CODE(MID($G2:$G$1000,8,1))-65,0))+1),26)+65)

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See J2 on sheet Data.
It is not as complicated as it looks, see below cells on the sheet.
But when assembled into one formula it gets long, because the calculation for the max value is done 4 times.
Sequential-lettering.xlsx
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Michael FowlerSolutions ConsultantCommented:
Here is an update.
LetterIncremented takes a range
Determines the largest value in the range
Then returns the next number with leading "A"s stripped.

It has been done by treating the string as a Base 26 number with "A" as zero.

Private Function Base26Max(rng As Range) As String
    Dim r As Range
    Dim i As Long, j As Long, currMax As Long
    Dim arr(1 To 4) As Integer
    Dim str As String
    
    currMax = 0
        
    For Each r In rng
        str = Right(r.Value, 4)
        j = 0
        For i = 1 To 4
            j = j + ((Asc(UCase(Mid(str, i, 1))) - 65) * (26 ^ (i - 1)))
        Next
        If j > currMax Then Base26Max = str
    Next
    
End Function

Function LetterIncremented(rng As Range) As String
    Dim arr(1 To 4) As Integer
    Dim i As Integer
    Dim str As String
    
    str = Base26Max(rng)
    
    For i = 1 To 4
        arr(i) = Asc(UCase(Mid(str, i, 1)))
    Next
    
    arr(4) = arr(4) + 1
    
    For i = 4 To 1 Step -1
        If (arr(i) > 90) Then
            arr(i) = 65
            arr(i - 1) = arr(i - 1) + 1
        End If
    Next
    
    If arr(1) = 65 Then
        If arr(2) = 65 Then
            If arr(3) = 65 Then
                LetterIncremented = Chr(arr(4))
            Else
                LetterIncremented = Chr(arr(3)) & Chr(arr(4))
            End If
        Else
            LetterIncremented = Chr(arr(2)) & Chr(arr(3)) & Chr(arr(4))
        End If
    Else
        LetterIncremented = Chr(arr(1)) & Chr(arr(2)) & Chr(arr(3)) & Chr(arr(4))
    End If
    
End Function

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Rob HensonFinance AnalystAuthor Commented:
Ejgil, the array formula works great. I have been able to successfully transfer it to the proper file with a couple of amendments to that file and it works well.

Michael74 - unfortunately I couldn't get the Function method to work and I didn't really want to go down the VBA route, issues with converting to xlsm and other users not being able to view from server or sharepoint sites etc.

Thanks to both
Rob H
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Michael FowlerSolutions ConsultantCommented:
All good Rob. Wasn't sure if it could be done with a formula and it sounded like it would be fun to do in VBA. I am glad Ejgil could give you the solution you needed
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