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1 | 1 |
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| 2 | +/* * |
| 3 | +* The time complexity of KMP algorithm is O(n) in the worst case |
| 4 | +* Example use case: Pattern = AABCAB Text = AAABACABAABCABAABCA |
| 5 | +* LPSArray = [ 0, 0, 1, 2, 3, 0 ] |
| 6 | +* Found = true, at position 13 |
| 7 | +* */ |
| 8 | + |
| 9 | +// Longest prefix suffix - generate an array of the lps for each pattern array value |
| 10 | +const createLPS = (pattern, patternLength) => { |
| 11 | + // initialise the current longest prefix suffix length and iterator index values |
| 12 | + const lps = [patternLength]; |
| 13 | + lps[0] = 0; |
| 14 | + |
| 15 | + let length = 0; |
| 16 | + let i = 1; |
| 17 | + // while there is still pattern to iterate over - calculate the lps for i = 1 to patternLength - 1 |
| 18 | + while (i < patternLength) { |
2 | 19 | /* *
|
3 | | - * The time complexity of KMP algorithm is O(n) in the worst case |
4 | | - * Example use case: Pattern = AABCAB Text = AAABACABAABCABAABCA |
5 | | - * LPSArray = [ 0, 0, 1, 2, 3, 0 ] |
6 | | - * Found = true, at position 13 |
7 | | - * */ |
8 | | - |
9 | | - // Longest prefix suffix - generate an array of the longest previous suffix for each pattern array value |
10 | | - const createLPS = (pattern, patternLength, lps) => { |
11 | | - // initialise the current longest prefix suffix length and iterator index values |
12 | | - lps[0] = 0; |
13 | | - let length = 0; |
14 | | - let i = 1; |
15 | | - // while there is still pattern to iterate over - calculate the lps for i = 1 to patternLength - 1 |
16 | | - while (i < patternLength) { |
17 | | - /* * |
18 | | - * if the pattern character at position i matches the pattern character at position length, then increment length, update |
19 | | - * the lps to the incremted length value and iterate to the next index i. |
20 | | - * */ |
21 | | - if (pattern.charAt(i) === pattern.charAt(length)) { |
22 | | - length++; |
23 | | - lps[i] = length; |
24 | | - i++; |
25 | | - } |
26 | | - // if a match is not found |
27 | | - else { |
28 | | - // if the length value is not 0, then set the length to be the lps value of index length - 1 |
29 | | - if (length !== 0) { |
30 | | - length = lps[length - 1]; |
31 | | - } |
32 | | - // else if length is 0, then set the lps at position i to length, i.e. 0 and increment i. |
33 | | - else { |
34 | | - lps[i] = length; |
35 | | - i++; |
36 | | - } |
37 | | - } |
38 | | - } |
39 | | - return lps; |
| 20 | + * if the pattern character at position i matches the pattern character at position length, |
| 21 | + * then increment length, update |
| 22 | + * the lps to the incremted length value and iterate to the next index i. |
| 23 | + * */ |
| 24 | + if (pattern.charAt(i) === pattern.charAt(length)) { |
| 25 | + length += 1; |
| 26 | + lps[i] = length; |
| 27 | + i += 1; |
| 28 | + // if not matching |
| 29 | + } else if (length !== 0) { |
| 30 | + // if the length value is not 0, then set the length to be the lps value of index length - 1 |
| 31 | + length = lps[length - 1]; |
| 32 | + } else { |
| 33 | + // else if length is 0, then set the lps at position i to length, i.e. 0 and increment i. |
| 34 | + lps[i] = length; |
| 35 | + i += 1; |
40 | 36 | }
|
| 37 | + } |
| 38 | + return lps; |
| 39 | +}; |
41 | 40 |
|
42 | | - /* * |
43 | | - * Invoke the Knuth-Morris-Pratt pattern matching algorithm to find a Pattern with a Text - this uses a precomputed prefix-suffix |
44 | | - * array/table to essentially skip chunks of the text that we know will match the pattern. |
45 | | - * This algorithm will return true if the pattern is a subset of the text, else it will return false. |
46 | | - * This algorithm accepts two strings, the pattern and text. |
47 | | - * The time complexity of the KMP algorithm is O(n) in the worst case. |
48 | | - * */ |
49 | | - const KMPSearch = (pattern, text) => { |
50 | | - const patternLength = pattern.length; // Often referred to as M |
51 | | - const textLength = text.length; // Often referred to as N |
| 41 | +/* * |
| 42 | +* Invoke the Knuth-Morris-Pratt pattern matching algorithm to find a Pattern with a Text - this |
| 43 | +* uses a precomputed prefix-suffix array/table to essentially skip chunks of the text that we |
| 44 | +* know will match the pattern. This algorithm will return true if the pattern is a subset of |
| 45 | +* the text, else it will return false. |
| 46 | +* This algorithm accepts two strings, the pattern and text. |
| 47 | +* The time complexity of the KMP algorithm is O(n) in the worst case. |
| 48 | +* */ |
| 49 | +const KMPSearch = (pattern, text) => { |
| 50 | + const patternLength = pattern.length; // Often referred to as M |
| 51 | + const textLength = text.length; // Often referred to as N |
52 | 52 |
|
53 | | - letlps=[patternLength];// Longest Pattern Suffix - array containing the lps for all pattern value positions |
54 | | - lps = createLPS(pattern, patternLength,lps); // This is preprocessed - before the text is searched for the pattern. |
55 | | - // console.log({ lpsArray: lps }) |
| 53 | + // Longest Pattern Suffix - array containing the lps for all pattern value positions |
| 54 | + constlps = createLPS(pattern, patternLength); // This is preprocessed. |
| 55 | + // console.log({ lpsArray: lps }) |
56 | 56 |
|
57 | | - let patternIndex = 0; // Referred to as P |
58 | | - let textIndex = 0; // Referred to as T |
59 | | - let found = false; |
| 57 | + let patternIndex = 0; // Referred to as P |
| 58 | + let textIndex = 0; // Referred to as T |
| 59 | + let found = false; |
60 | 60 |
|
61 | | - // While there is still text left to iterate over and the pattern has not yet been found |
62 | | - while (textIndex < textLength && found === false) { |
63 | | - // if the pattern character at pattern index P equals the text character at text index T, then increment the text and pattern indexes |
64 | | - if (pattern.charAt(patternIndex) === text.charAt(textIndex)) { |
65 | | - textIndex++; |
66 | | - patternIndex++; |
67 | | - } |
68 | | - /* * |
69 | | - * if the pattern index equals the pattern length then the pattern has been successfully found, as such the pattern is a subset of |
70 | | - * the text the pattern index is set to the longest pattern suffix value (the index is decremented due to being zero indexed). |
71 | | - * */ |
72 | | - if(patternIndex===patternLength){ |
73 | | - // console.log(`Pattern found at index ${textIndex-patternIndex}`); |
74 | | - patternIndex=lps[patternIndex-1]; |
75 | | - found=true; |
76 | | - } |
77 | | - /* * |
78 | | - * else if there is still text left to iterate over and the pattern character does not match the text character at their respective |
79 | | - * index positions, then check of the pattern Index is 0, i.e. if it is the first pattern position. If so then jump to the next text |
80 | | - * character, else (this is not the first pattern position), then update the pattern index using the generated longest pattern suffix, |
81 | | - * to skip ahead of matching values. This logic will only be encountered after T number of mismatches. |
82 | | - * */ |
83 | | - elseif(textIndex<textLength&&pattern.charAt(patternIndex)!==text.charAt(textIndex)){ |
84 | | - if(patternIndex===0) |
85 | | - textIndex=textIndex+1; |
86 | | - else |
87 | | - patternIndex = lps[patternIndex - 1]; |
88 | | - } |
89 | | - } |
90 | | - // Pattern has not been found, return false. Else return true. |
91 | | - if (!found) { |
92 | | - // console.log('The pattern was not found!') |
93 | | - return false |
94 | | - } |
95 | | - return true |
96 | | -}; |
| 61 | + // While there is still text left to iterate over and the pattern has not yet been found |
| 62 | + while (textIndex < textLength && found === false) { |
| 63 | + // if the pattern char at index pos P equals the text char at text pos T, then increment indexes |
| 64 | + if (pattern.charAt(patternIndex) === text.charAt(textIndex)) { |
| 65 | + textIndex+=1; |
| 66 | + patternIndex+=1; |
| 67 | + } |
| 68 | + /* * |
| 69 | + * if the pattern index equals the pattern length then the pattern has been successfully |
| 70 | + * found, as such the pattern is a subset of the text the pattern index is set to the longest |
| 71 | + * pattern suffix value (the index is decremented due to being zero indexed). |
| 72 | + * */ |
| 73 | + if(patternIndex===patternLength){ |
| 74 | + // console.log(`Pattern found at index ${textIndex-patternIndex}`); |
| 75 | + patternIndex=lps[patternIndex-1]; |
| 76 | + found=true; |
| 77 | + }elseif(textIndex<textLength&&pattern.charAt(patternIndex)!==text.charAt(textIndex)){ |
| 78 | + /* * |
| 79 | + * else if there is still text left to iterate over and the pattern character does not match |
| 80 | + * the text characterat their respective index positions, then check of the pattern Index is 0, |
| 81 | + * i.e. if it is the first pattern position. If so then jump to the next text character, else |
| 82 | + * (this is not the first pattern position), then update the pattern index using the generated |
| 83 | + * longest prefix suffix, to skip ahead of matching values. This logic will only be encountered |
| 84 | + * after T number of mismatches. |
| 85 | + * */ |
| 86 | + if(patternIndex===0)textIndex+=1; |
| 87 | + elsepatternIndex = lps[patternIndex - 1]; |
| 88 | + } |
| 89 | + } |
| 90 | + // Pattern has not been found, return false. Else return true. |
| 91 | + if (!found) { |
| 92 | + // console.log('The pattern was not found!') |
| 93 | + return false; |
| 94 | + } |
| 95 | + return true; |
| 96 | +}; |
97 | 97 |
|
98 | | -module.exports = { |
99 | | - KMPSearch |
100 | | -}; |
| 98 | +module.exports = { |
| 99 | + KMPSearch, |
| 100 | +}; |
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