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reduce OSADamerau–Levenshtein space complexity from O(m*n) to O(3 * min(m*n)) #17

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merged 1 commit into from Jul 3, 2022
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22 changes: 11 additions & 11 deletions levenshtein.go
Expand Up @@ -65,40 +65,40 @@ func OSADamerauLevenshteinDistance(str1, str2 string) int {
return runeStr1len
} else if utils.Equal(runeStr1, runeStr2) {
return 0
} else if runeStr1len < runeStr2len {
return OSADamerauLevenshteinDistance(str2, str1)
}

// 2D Array
matrix := make([][]int, runeStr1len+1)
for i := 0; i <= runeStr1len; i++ {
row := utils.Min(runeStr1len+1, 3)
matrix := make([][]int, row)
for i := 0; i < row; i++ {
matrix[i] = make([]int, runeStr2len+1)
for j := 0; j <= runeStr2len; j++ {
matrix[i][j] = 0
}
}

for i := 0; i <= runeStr1len; i++ {
matrix[i][0] = i
}

for j := 0; j <= runeStr2len; j++ {
matrix[0][j] = j
}

var count int
for i := 1; i <= runeStr1len; i++ {
matrix[i%3][0] = i
for j := 1; j <= runeStr2len; j++ {
if runeStr1[i-1] == runeStr2[j-1] {
count = 0
} else {
count = 1
}

matrix[i][j] = utils.Min(utils.Min(matrix[i-1][j]+1, matrix[i][j-1]+1), matrix[i-1][j-1]+count) // insertion, deletion, substitution
matrix[i%3][j] = utils.Min(utils.Min(matrix[(i-1)%3][j]+1, matrix[i%3][j-1]+1),
matrix[(i-1)%3][j-1]+count) // insertion, deletion, substitution
if i > 1 && j > 1 && runeStr1[i-1] == runeStr2[j-2] && runeStr1[i-2] == runeStr2[j-1] {
matrix[i][j] = utils.Min(matrix[i][j], matrix[i-2][j-2]+1) // translation
matrix[i%3][j] = utils.Min(matrix[i%3][j], matrix[(i-2)%3][j-2]+1) // translation
}
}
}
return matrix[runeStr1len][runeStr2len]
return matrix[runeStr1len%3][runeStr2len]
}

// DamerauLevenshteinDistance calculate the distance between two string
Expand Down