All Downloads are FREE. Search and download functionalities are using the official Maven repository.

com.groupbyinc.common.apache.commons.text.similarity.LongestCommonSubsequence Maven / Gradle / Ivy

There is a newer version: 198
Show newest version
/*
 * Licensed to the Apache Software Foundation (ASF) under one or more
 * contributor license agreements.  See the NOTICE file distributed with
 * this work for additional information regarding copyright ownership.
 * The ASF licenses this file to You under the Apache License, Version 2.0
 * (the "License"); you may not use this file except in compliance with
 * the License.  You may obtain a copy of the License at
 *
 *      http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */
package org.apache.commons.text.similarity;

/**
 * A similarity algorithm indicating the length of the longest common subsequence between two strings.
 *
 * 

* The Longest common subsequence algorithm returns the length of the longest subsequence that two strings have in * common. Two strings that are entirely different, return a value of 0, and two strings that return a value * of the commonly shared length implies that the strings are completely the same in value and position. * Note. Generally this algorithm is fairly inefficient, as for length m, n of the input * CharSequence's left and right respectively, the runtime of the * algorithm is O(m*n). *

* *

* This implementation is based on the Longest Commons Substring algorithm * from * https://en.wikipedia.org/wiki/Longest_common_subsequence_problem. *

* *

For further reading see:

* *

Lothaire, M. Applied combinatorics on words. New York: Cambridge U Press, 2005. 12-13

* * @since 1.0 */ public class LongestCommonSubsequence implements SimilarityScore { /** * Calculates longest common subsequence similarity score of two CharSequence's passed as * input. * * @param left first character sequence * @param right second character sequence * @return longestCommonSubsequenceLength * @throws IllegalArgumentException * if either String input {@code null} */ @Override public Integer apply(final CharSequence left, final CharSequence right) { // Quick return for invalid inputs if (left == null || right == null) { throw new IllegalArgumentException("Inputs must not be null"); } return longestCommonSubsequence(left, right).length(); } /** * Computes the longest common subsequence between the two CharSequence's passed as input. * *

* Note, a substring and subsequence are not necessarily the same thing. Indeed, abcxyzqrs and * xyzghfm have both the same common substring and subsequence, namely xyz. However, * axbyczqrs and abcxyzqtv have the longest common subsequence xyzq because a * subsequence need not have adjacent characters. *

* *

* For reference, we give the definition of a subsequence for the reader: a subsequence is a sequence that * can be derived from another sequence by deleting some elements without changing the order of the remaining * elements. *

* * @param left first character sequence * @param right second character sequence * @return the longest common subsequence found * @throws IllegalArgumentException * if either String input {@code null} * @deprecated Deprecated as of 1.2 due to a typo in the method name. * Use {@link #longestCommonSubsequence(CharSequence, CharSequence)} instead. * This method will be removed in 2.0. */ @Deprecated public CharSequence logestCommonSubsequence(final CharSequence left, final CharSequence right) { return longestCommonSubsequence(left, right); } /** * Computes the longest common subsequence between the two CharSequence's passed as * input. * *

* Note, a substring and subsequence are not necessarily the same thing. Indeed, abcxyzqrs and * xyzghfm have both the same common substring and subsequence, namely xyz. However, * axbyczqrs and abcxyzqtv have the longest common subsequence xyzq because a * subsequence need not have adjacent characters. *

* *

* For reference, we give the definition of a subsequence for the reader: a subsequence is a sequence that * can be derived from another sequence by deleting some elements without changing the order of the remaining * elements. *

* * @param left first character sequence * @param right second character sequence * @return the longest common subsequence found * @throws IllegalArgumentException * if either String input {@code null} * @since 1.2 */ public CharSequence longestCommonSubsequence(final CharSequence left, final CharSequence right) { // Quick return if (left == null || right == null) { throw new IllegalArgumentException("Inputs must not be null"); } final StringBuilder longestCommonSubstringArray = new StringBuilder(Math.max(left.length(), right.length())); final int[][] lcsLengthArray = longestCommonSubstringLengthArray(left, right); int i = left.length() - 1; int j = right.length() - 1; int k = lcsLengthArray[left.length()][right.length()] - 1; while (k >= 0) { if (left.charAt(i) == right.charAt(j)) { longestCommonSubstringArray.append(left.charAt(i)); i = i - 1; j = j - 1; k = k - 1; } else if (lcsLengthArray[i + 1][j] < lcsLengthArray[i][j + 1]) { i = i - 1; } else { j = j - 1; } } return longestCommonSubstringArray.reverse().toString(); } /** * * Computes the lcsLengthArray for the sake of doing the actual lcs calculation. This is the * dynamic programming portion of the algorithm, and is the reason for the runtime complexity being * O(m*n), where m=left.length() and n=right.length(). * * @param left first character sequence * @param right second character sequence * @return lcsLengthArray */ public int[][] longestCommonSubstringLengthArray(final CharSequence left, final CharSequence right) { final int[][] lcsLengthArray = new int[left.length() + 1][right.length() + 1]; for (int i = 0; i < left.length(); i++) { for (int j = 0; j < right.length(); j++) { if (i == 0) { lcsLengthArray[i][j] = 0; } if (j == 0) { lcsLengthArray[i][j] = 0; } if (left.charAt(i) == right.charAt(j)) { lcsLengthArray[i + 1][j + 1] = lcsLengthArray[i][j] + 1; } else { lcsLengthArray[i + 1][j + 1] = Math.max(lcsLengthArray[i + 1][j], lcsLengthArray[i][j + 1]); } } } return lcsLengthArray; } }




© 2015 - 2024 Weber Informatics LLC | Privacy Policy