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// SPDX-License-Identifier: GPL-2.0
#include "levenshtein.h"
#include <errno.h>
#include <stdlib.h>
#include <string.h>
/*
* This function implements the Damerau-Levenshtein algorithm to
* calculate a distance between strings.
*
* Basically, it says how many letters need to be swapped, substituted,
* deleted from, or added to string1, at least, to get string2.
*
* The idea is to build a distance matrix for the substrings of both
* strings. To avoid a large space complexity, only the last three rows
* are kept in memory (if swaps had the same or higher cost as one deletion
* plus one insertion, only two rows would be needed).
*
* At any stage, "i + 1" denotes the length of the current substring of
* string1 that the distance is calculated for.
*
* row2 holds the current row, row1 the previous row (i.e. for the substring
* of string1 of length "i"), and row0 the row before that.
*
* In other words, at the start of the big loop, row2[j + 1] contains the
* Damerau-Levenshtein distance between the substring of string1 of length
* "i" and the substring of string2 of length "j + 1".
*
* All the big loop does is determine the partial minimum-cost paths.
*
* It does so by calculating the costs of the path ending in characters
* i (in string1) and j (in string2), respectively, given that the last
* operation is a substitution, a swap, a deletion, or an insertion.
*
* This implementation allows the costs to be weighted:
*
* - w (as in "sWap")
* - s (as in "Substitution")
* - a (for insertion, AKA "Add")
* - d (as in "Deletion")
*
* Note that this algorithm calculates a distance _iff_ d == a.
*/
int levenshtein(const char *string1, const char *string2,
int w, int s, int a, int d)
{
int len1 = strlen(string1), len2 = strlen(string2);
int *row0 = malloc(sizeof(int) * (len2 + 1));
int *row1 = malloc(sizeof(int) * (len2 + 1));
int *row2 = malloc(sizeof(int) * (len2 + 1));
int i, j;
for (j = 0; j <= len2; j++)
row1[j] = j * a;
for (i = 0; i < len1; i++) {
int *dummy;
row2[0] = (i + 1) * d;
for (j = 0; j < len2; j++) {
/* substitution */
row2[j + 1] = row1[j] + s * (string1[i] != string2[j]);
/* swap */
if (i > 0 && j > 0 && string1[i - 1] == string2[j] &&
string1[i] == string2[j - 1] &&
row2[j + 1] > row0[j - 1] + w)
row2[j + 1] = row0[j - 1] + w;
/* deletion */
if (row2[j + 1] > row1[j + 1] + d)
row2[j + 1] = row1[j + 1] + d;
/* insertion */
if (row2[j + 1] > row2[j] + a)
row2[j + 1] = row2[j] + a;
}
dummy = row0;
row0 = row1;
row1 = row2;
row2 = dummy;
}
i = row1[len2];
free(row0);
free(row1);
free(row2);
return i;
}
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