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Add LCS.
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* [Knuth–Morris–Pratt Algorithm](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/knuth-morris-pratt) - substring search
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* [Rabin Karp Algorithm](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/rabin-karp) - substring search
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* [Longest Common Subsequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/longest-common-subsequnce) (LCS)
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* longest common substring
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* [Longest Common Substring](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/longest-common-substring)
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* **Search**
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* [Binary Search](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/search/binary-search)
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* **Sorting**
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* **Dynamic Programming**
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* [Levenshtein Distance](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/levenshtein-distance) - minimum edit distance between two sequences
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* [Longest Common Subsequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/longest-common-subsequnce) (LCS)
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* [Longest Common Substring](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/longest-common-substring)
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* Increasing subsequence
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* Knapsack problem
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* Maximum subarray
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src/algorithms/string/longest-common-substring/README.md
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src/algorithms/string/longest-common-substring/README.md
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# Longest Common Substring Problem
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The longest common substring problem is to find the longest string
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(or strings) that is a substring (or are substrings) of two or more
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strings.
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## Example
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The longest common substring of the strings `ABABC`, `BABCA` and
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`ABCBA` is string `ABC` of length 3. Other common substrings are
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`A`, `AB`, `B`, `BA`, `BC` and `C`.
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```
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ABABC
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BABCA
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ABCBA
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```
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## References
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- [Wikipedia](https://en.wikipedia.org/wiki/Longest_common_substring_problem)
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- [YouTube](https://www.youtube.com/watch?v=BysNXJHzCEs)
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import longestCommonSubstring from '../longestCommonSubstring';
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describe('longestCommonSubstring', () => {
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it('should find longest common substring between two strings', () => {
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expect(longestCommonSubstring('', '')).toBe('');
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expect(longestCommonSubstring('ABC', '')).toBe('');
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expect(longestCommonSubstring('', 'ABC')).toBe('');
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expect(longestCommonSubstring('ABABC', 'BABCA')).toBe('BABC');
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expect(longestCommonSubstring('BABCA', 'ABCBA')).toBe('ABC');
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});
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});
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/**
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* @param {string} s1
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* @param {string} s2
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* @return {string}
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*/
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export default function longestCommonSubstring(s1, s2) {
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// Init the matrix of all substring lengths to use Dynamic Programming approach.
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const substringMatrix = Array(s2.length + 1).fill(null).map(() => {
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return Array(s1.length + 1).fill(null);
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});
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// Fill the first row and first column with zeros to provide initial values.
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for (let columnIndex = 0; columnIndex <= s1.length; columnIndex += 1) {
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substringMatrix[0][columnIndex] = 0;
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}
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for (let rowIndex = 0; rowIndex <= s2.length; rowIndex += 1) {
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substringMatrix[rowIndex][0] = 0;
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}
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// Build the matrix of all substring lengths to use Dynamic Programming approach.
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let longestSubstringLength = 0;
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let longestSubstringColumn = 0;
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let longestSubstringRow = 0;
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for (let rowIndex = 1; rowIndex <= s2.length; rowIndex += 1) {
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for (let columnIndex = 1; columnIndex <= s1.length; columnIndex += 1) {
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if (s1[columnIndex - 1] === s2[rowIndex - 1]) {
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substringMatrix[rowIndex][columnIndex] = substringMatrix[rowIndex - 1][columnIndex - 1] + 1;
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} else {
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substringMatrix[rowIndex][columnIndex] = 0;
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}
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// Try to find the biggest length of all common substring lengths
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// and to memorize its last character position (indices)
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if (substringMatrix[rowIndex][columnIndex] > longestSubstringLength) {
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longestSubstringLength = substringMatrix[rowIndex][columnIndex];
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longestSubstringColumn = columnIndex;
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longestSubstringRow = rowIndex;
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}
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}
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}
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if (longestSubstringLength === 0) {
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// Longest common substring has not been found.
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return '';
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}
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// Detect the longest substring from the matrix.
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let longestSubstring = '';
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while (substringMatrix[longestSubstringRow][longestSubstringColumn] > 0) {
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longestSubstring = s1[longestSubstringColumn - 1] + longestSubstring;
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longestSubstringRow -= 1;
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longestSubstringColumn -= 1;
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}
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return longestSubstring;
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}
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