    001/*
    002 * Licensed to the Apache Software Foundation (ASF) under one or more
    003 * contributor license agreements.  See the NOTICE file distributed with
    004 * this work for additional information regarding copyright ownership.
    005 * The ASF licenses this file to You under the Apache License, Version 2.0
    006 * (the "License"); you may not use this file except in compliance with
    007 * the License.  You may obtain a copy of the License at
    008 *
    009 *      http://www.apache.org/licenses/LICENSE-2.0
    010 *
    011 * Unless required by applicable law or agreed to in writing, software
    012 * distributed under the License is distributed on an "AS IS" BASIS,
    013 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
    014 * See the License for the specific language governing permissions and
    015 * limitations under the License.
    016 */
    017package org.apache.commons.text.similarity;
    018
    019/**
    020 * A similarity algorithm indicating the length of the longest common subsequence between two strings.
    021 *
    022 * <p>
    023 * The Longest common subsequence algorithm returns the length of the longest subsequence that two strings have in
    024 * common. Two strings that are entirely different, return a value of 0, and two strings that return a value
    025 * of the commonly shared length implies that the strings are completely the same in value and position.
    026 * <i>Note.</i>  Generally this algorithm is fairly inefficient, as for length <i>m</i>, <i>n</i> of the input
    027 * <code>CharSequence</code>'s <code>left</code> and <code>right</code> respectively, the runtime of the
    028 * algorithm is <i>O(m*n)</i>.
    029 * </p>
    030 *
    031 * <p>
    032 * This implementation is based on the Longest Commons Substring algorithm
    033 * from <a href="https://en.wikipedia.org/wiki/Longest_common_subsequence_problem">
    034 * https://en.wikipedia.org/wiki/Longest_common_subsequence_problem</a>.
    035 * </p>
    036 *
    037 * <p>For further reading see:</p>
    038 *
    039 * <p>Lothaire, M. <i>Applied combinatorics on words</i>. New York: Cambridge U Press, 2005. <b>12-13</b></p>
    040 *
    041 * @since 1.0
    042 */
    043public class LongestCommonSubsequence implements SimilarityScore<Integer> {
    044
    045    /**
    046     * Calculates longest common subsequence similarity score of two <code>CharSequence</code>'s passed as
    047     * input.
    048     *
    049     * @param left first character sequence
    050     * @param right second character sequence
    051     * @return longestCommonSubsequenceLength
    052     * @throws IllegalArgumentException
    053     *             if either String input {@code null}
    054     */
    055    @Override
    056    public Integer apply(final CharSequence left, final CharSequence right) {
    057        // Quick return for invalid inputs
    058        if (left == null || right == null) {
    059            throw new IllegalArgumentException("Inputs must not be null");
    060        }
    061        return longestCommonSubsequence(left, right).length();
    062    }
    063
    064    /**
    065     * Computes the longest common subsequence between the two <code>CharSequence</code>'s passed as input.
    066     *
    067     * <p>
    068     * Note, a substring and subsequence are not necessarily the same thing. Indeed, <code>abcxyzqrs</code> and
    069     * <code>xyzghfm</code> have both the same common substring and subsequence, namely <code>xyz</code>. However,
    070     * <code>axbyczqrs</code> and <code>abcxyzqtv</code> have the longest common subsequence <code>xyzq</code> because a
    071     * subsequence need not have adjacent characters.
    072     * </p>
    073     *
    074     * <p>
    075     * For reference, we give the definition of a subsequence for the reader: a <i>subsequence</i> is a sequence that
    076     * can be derived from another sequence by deleting some elements without changing the order of the remaining
    077     * elements.
    078     * </p>
    079     *
    080     * @param left first character sequence
    081     * @param right second character sequence
    082     * @return the longest common subsequence found
    083     * @throws IllegalArgumentException
    084     *             if either String input {@code null}
    085     * @deprecated Deprecated as of 1.2 due to a typo in the method name.
    086     *              Use {@link #longestCommonSubsequence(CharSequence, CharSequence)} instead.
    087     *              This method will be removed in 2.0.
    088     */
    089    @Deprecated
    090    public CharSequence logestCommonSubsequence(final CharSequence left, final CharSequence right) {
    091        return longestCommonSubsequence(left, right);
    092    }
    093
    094   /**
    095    * Computes the longest common subsequence between the two <code>CharSequence</code>'s passed as
    096    * input.
    097    *
    098    * <p>
    099    * Note, a substring and subsequence are not necessarily the same thing. Indeed, <code>abcxyzqrs</code> and
    100    * <code>xyzghfm</code> have both the same common substring and subsequence, namely <code>xyz</code>. However,
    101    * <code>axbyczqrs</code> and <code>abcxyzqtv</code> have the longest common subsequence <code>xyzq</code> because a
    102    * subsequence need not have adjacent characters.
    103    * </p>
    104    *
    105    * <p>
    106    * For reference, we give the definition of a subsequence for the reader: a <i>subsequence</i> is a sequence that
    107    * can be derived from another sequence by deleting some elements without changing the order of the remaining
    108    * elements.
    109    * </p>
    110    *
    111    * @param left first character sequence
    112    * @param right second character sequence
    113    * @return the longest common subsequence found
    114    * @throws IllegalArgumentException
    115    *             if either String input {@code null}
    116    * @since 1.2
    117    */
    118   public CharSequence longestCommonSubsequence(final CharSequence left, final CharSequence right) {
    119       // Quick return
    120       if (left == null || right == null) {
    121           throw new IllegalArgumentException("Inputs must not be null");
    122       }
    123       final StringBuilder longestCommonSubstringArray = new StringBuilder(Math.max(left.length(), right.length()));
    124       final int[][] lcsLengthArray = longestCommonSubstringLengthArray(left, right);
    125       int i = left.length() - 1;
    126       int j = right.length() - 1;
    127       int k = lcsLengthArray[left.length()][right.length()] - 1;
    128       while (k >= 0) {
    129           if (left.charAt(i) == right.charAt(j)) {
    130               longestCommonSubstringArray.append(left.charAt(i));
    131               i = i - 1;
    132               j = j - 1;
    133               k = k - 1;
    134           } else if (lcsLengthArray[i + 1][j] < lcsLengthArray[i][j + 1]) {
    135               i = i - 1;
    136           } else {
    137               j = j - 1;
    138           }
    139       }
    140       return longestCommonSubstringArray.reverse().toString();
    141   }
    142
    143    /**
    144     *
    145     * Computes the lcsLengthArray for the sake of doing the actual lcs calculation. This is the
    146     * dynamic programming portion of the algorithm, and is the reason for the runtime complexity being
    147     * O(m*n), where m=left.length() and n=right.length().
    148     *
    149     * @param left first character sequence
    150     * @param right second character sequence
    151     * @return lcsLengthArray
    152     */
    153    public int[][] longestCommonSubstringLengthArray(final CharSequence left, final CharSequence right) {
    154        final int[][] lcsLengthArray = new int[left.length() + 1][right.length() + 1];
    155        for (int i = 0; i < left.length(); i++) {
    156            for (int j = 0; j < right.length(); j++) {
    157                if (i == 0) {
    158                    lcsLengthArray[i][j] = 0;
    159                }
    160                if (j == 0) {
    161                    lcsLengthArray[i][j] = 0;
    162                }
    163                if (left.charAt(i) == right.charAt(j)) {
    164                    lcsLengthArray[i + 1][j + 1] = lcsLengthArray[i][j] + 1;
    165                } else {
    166                    lcsLengthArray[i + 1][j + 1] = Math.max(lcsLengthArray[i + 1][j], lcsLengthArray[i][j + 1]);
    167                }
    168            }
    169        }
    170        return lcsLengthArray;
    171    }
    172
    173}