Most efficient way to calculate Levenshtein distance

Question

I just implemented a best match file search algorithm to find the closest match to a string in a dictionary. After profiling my code, I found out that the overwhelming major

Answer 1

I modified the Levenshtein distance VBA function found on this post to use a one dimensional array. It performs much faster.

'Calculate the Levenshtein Distance between two strings (the number of insertions,
'deletions, and substitutions needed to transform the first string into the second)

Public Function LevenshteinDistance2(ByRef s1 As String, ByRef s2 As String) As Long
Dim L1 As Long, L2 As Long, D() As Long, LD As Long 'Length of input strings and distance matrix
Dim i As Long, j As Long, ss2 As Long, ssL As Long, cost As Long 'loop counters, loop step, loop start, and cost of substitution for current letter
Dim cI As Long, cD As Long, cS As Long 'cost of next Insertion, Deletion and Substitution
Dim L1p1 As Long, L1p2 As Long 'Length of S1 + 1, Length of S1 + 2

L1 = Len(s1): L2 = Len(s2)
L1p1 = L1 + 1
L1p2 = L1 + 2
LD = (((L1 + 1) * (L2 + 1))) - 1
ReDim D(0 To LD)
ss2 = L1 + 1

For i = 0 To L1 Step 1: D(i) = i: Next i                'setup array positions 0,1,2,3,4,...
For j = 0 To LD Step ss2: D(j) = j / ss2: Next j        'setup array positions 0,1,2,3,4,...

For j = 1 To L2
    ssL = (L1 + 1) * j
    For i = (ssL + 1) To (ssL + L1)
        If Mid$(s1, i Mod ssL, 1) <> Mid$(s2, j, 1) Then cost = 1 Else cost = 0
        cI = D(i - 1) + 1
        cD = D(i - L1p1) + 1
        cS = D(i - L1p2) + cost

        If cI <= cD Then 'Insertion or Substitution
            If cI <= cS Then D(i) = cI Else D(i) = cS
        Else 'Deletion or Substitution
            If cD <= cS Then D(i) = cD Else D(i) = cS
        End If
    Next i
Next j

LevenshteinDistance2 = D(LD)
End Function

I have tested this function with string 's1' of length 11,304 and 's2' of length 5,665 ( > 64 million character comparisons). With the above single dimension version of the function, the execution time is ~24 seconds on my machine. The original two dimensional function that I referenced in the link above requires ~37 seconds for the same strings. I have optimized the single dimensional function further as shown below and it requires ~10 seconds for the same strings.

'Calculate the Levenshtein Distance between two strings (the number of insertions,
'deletions, and substitutions needed to transform the first string into the second)
Public Function LevenshteinDistance(ByRef s1 As String, ByRef s2 As String) As Long
Dim L1 As Long, L2 As Long, D() As Long, LD As Long         'Length of input strings and distance matrix
Dim i As Long, j As Long, ss2 As Long                       'loop counters, loop step
Dim ssL As Long, cost As Long                               'loop start, and cost of substitution for current letter
Dim cI As Long, cD As Long, cS As Long                      'cost of next Insertion, Deletion and Substitution
Dim L1p1 As Long, L1p2 As Long                              'Length of S1 + 1, Length of S1 + 2
Dim sss1() As String, sss2() As String                      'Character arrays for string S1 & S2

L1 = Len(s1): L2 = Len(s2)
L1p1 = L1 + 1
L1p2 = L1 + 2
LD = (((L1 + 1) * (L2 + 1))) - 1
ReDim D(0 To LD)
ss2 = L1 + 1

For i = 0 To L1 Step 1: D(i) = i: Next i                    'setup array positions 0,1,2,3,4,...
For j = 0 To LD Step ss2: D(j) = j / ss2: Next j            'setup array positions 0,1,2,3,4,...

ReDim sss1(1 To L1)                                         'Size character array S1
ReDim sss2(1 To L2)                                         'Size character array S2
For i = 1 To L1 Step 1: sss1(i) = Mid$(s1, i, 1): Next i    'Fill S1 character array
For i = 1 To L2 Step 1: sss2(i) = Mid$(s2, i, 1): Next i    'Fill S2 character array

For j = 1 To L2
    ssL = (L1 + 1) * j
    For i = (ssL + 1) To (ssL + L1)
        If sss1(i Mod ssL) <> sss2(j) Then cost = 1 Else cost = 0
        cI = D(i - 1) + 1
        cD = D(i - L1p1) + 1
        cS = D(i - L1p2) + cost
        If cI <= cD Then 'Insertion or Substitution
            If cI <= cS Then D(i) = cI Else D(i) = cS
        Else 'Deletion or Substitution
            If cD <= cS Then D(i) = cD Else D(i) = cS
        End If
    Next i
Next j

LevenshteinDistance = D(LD)
End Function