levenSubstringIC - find leven-closest substring in larger string. complexity O(m*n) [1023862]

< > BotCompany Repo | #1023862 // levenSubstringIC - find leven-closest substring in larger string. complexity O(m*n)

JavaX fragment (include) [tags: use-pretranspiled]

// Calculates the fuzzy match of needle in haystack,
// using a modified version of the Levenshtein distance algorithm.
// ported from: http://ginstrom.com/scribbles/2007/12/01/fuzzy-substring-matching-with-levenshtein-distance-in-python/

static int levenSubstringIC(S haystack, S needle) {
  int m = strL(needle), n = strL(haystack);

  // base cases
  if (containsIgnoreCase(haystack, needle)) ret 0;

  int[] v0 = new int[n+1], v1 = new int[n+1];
  for i to m: { // iterate over needle
    v1[0] = i+1;
    for j to n: { // iterate over haystack
      int cost = neqic(needle.charAt(i), haystack.charAt(j)) ? 1 : 0;
      v1[j+1] = min3(v1[j]+1,       // insertion
                       v0[j+1]+1,   // deletion
                       v0[j]+cost); // substitution
    }
    int[] temp = v0;
    v0 = v1;
    v1 = temp;
  }
  ret minOfIntArray(v0);
}

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Snippet ID:

#1023862

Snippet name:

levenSubstringIC - find leven-closest substring in larger string. complexity O(m*n)

Eternal ID of this version:

#1023862/18

Text MD5:

9b4b6ac905b2e9f6c10602294e1a53f2

Transpilation MD5:

af3329563c64aeeca709f9d60c287c34

Author:

stefan

Category:

javax / a.i.

Type:

JavaX fragment (include)

Public (visible to everyone):

Yes

Archived (hidden from active list):

Created/modified:

2019-11-11 14:59:59

Source code size:

878 bytes / 25 lines

Pitched / IR pitched:

No / No

Views / Downloads:

203 / 355

Version history:

17 change(s)

Referenced in:

[show references]

1	// Calculates the fuzzy match of needle in haystack,
2	// using a modified version of the Levenshtein distance algorithm.
3	// ported from: http://ginstrom.com/scribbles/2007/12/01/fuzzy-substring-matching-with-levenshtein-distance-in-python/
4
5	static int levenSubstringIC(S haystack, S needle) {
6	int m = strL(needle), n = strL(haystack);
7
8	// base cases
9	if (containsIgnoreCase(haystack, needle)) ret 0;
10
11	int[] v0 = new int[n+1], v1 = new int[n+1];
12	for i to m: { // iterate over needle
13	v1[0] = i+1;
14	for j to n: { // iterate over haystack
15	int cost = neqic(needle.charAt(i), haystack.charAt(j)) ? 1 : 0;
16	v1[j+1] = min3(v1[j]+1, // insertion
17	v0[j+1]+1, // deletion
18	v0[j]+cost); // substitution
19	}
20	int[] temp = v0;
21	v0 = v1;
22	v1 = temp;
23	}
24	ret minOfIntArray(v0);
25	}