    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}'s {@code left} and {@code right} respectively, the runtime of the
    028 * algorithm is <i>O(m*n)</i>.
    029 * </p>
    030 *
    031 * <p>
    032 * As of version 1.10, a more space-efficient of the algorithm is implemented. The new algorithm has linear space
    033 * complexity instead of quadratic. However, time complexity is still quadratic in the size of input strings.
    034 * </p>
    035 *
    036 * <p>
    037 * The implementation is based on Hirschberg's Longest Commons Substring algorithm (cited below).
    038 * </p>
    039 *
    040 * <p>For further reading see:</p>
    041 * <ul>
    042 * <li>
    043 * Lothaire, M. <i>Applied combinatorics on words</i>. New York: Cambridge U Press, 2005. <b>12-13</b>
    044 * </li>
    045 * <li>
    046 * D. S. Hirschberg, "A linear space algorithm for computing maximal common subsequences," CACM, 1975, pp. 341--343.
    047 * </li>
    048 * </ul>
    049 *
    050 * @since 1.0
    051 */
    052public class LongestCommonSubsequence implements SimilarityScore<Integer> {
    053
    054    /**
    055     * Singleton instance.
    056     */
    057    static final LongestCommonSubsequence INSTANCE = new LongestCommonSubsequence();
    058
    059    /**
    060     * An implementation of "ALG B" from Hirschberg's CACM '71 paper.
    061     * Assuming the first input sequence is of size <code>m</code> and the second input sequence is of size
    062     * <code>n</code>, this method returns the last row of the dynamic programming (DP) table when calculating
    063     * the LCS of the two sequences in <i>O(m*n)</i> time and <i>O(n)</i> space.
    064     * The last element of the returned array, is the size of the LCS of the two input sequences.
    065     *
    066     * @param left first input sequence.
    067     * @param right second input sequence.
    068     * @return last row of the dynamic-programming (DP) table for calculating the LCS of <code>left</code> and <code>right</code>
    069     * @since 1.10.0
    070     */
    071    private static int[] algorithmB(final CharSequence left, final CharSequence right) {
    072        final int m = left.length();
    073        final int n = right.length();
    074
    075        // Creating an array for storing two rows of DP table
    076        final int[][] dpRows = new int[2][1 + n];
    077
    078        for (int i = 1; i <= m; i++) {
    079            // K(0, j) <- K(1, j) [j = 0...n], as per the paper:
    080            // Since we have references in Java, we can swap references instead of literal copying.
    081            // We could also use a "binary index" using modulus operator, but directly swapping the
    082            // two rows helps readability and keeps the code consistent with the algorithm description
    083            // in the paper.
    084            final int[] temp = dpRows[0];
    085            dpRows[0] = dpRows[1];
    086            dpRows[1] = temp;
    087
    088            for (int j = 1; j <= n; j++) {
    089                if (left.charAt(i - 1) == right.charAt(j - 1)) {
    090                    dpRows[1][j] = dpRows[0][j - 1] + 1;
    091                } else {
    092                    dpRows[1][j] = Math.max(dpRows[1][j - 1], dpRows[0][j]);
    093                }
    094            }
    095        }
    096
    097        // LL(j) <- K(1, j) [j=0...n], as per the paper:
    098        // We don't need literal copying of the array, we can just return the reference
    099        return dpRows[1];
    100    }
    101
    102    /**
    103     * An implementation of "ALG C" from Hirschberg's CACM '71 paper.
    104     * Assuming the first input sequence is of size <code>m</code> and the second input sequence is of size
    105     * <code>n</code>, this method returns the Longest Common Subsequence (LCS) of the two sequences in
    106     * <i>O(m*n)</i> time and <i>O(m+n)</i> space.
    107     *
    108     * @param left first input sequence.
    109     * @param right second input sequence.
    110     * @return the LCS of <code>left</code> and <code>right</code>
    111     * @since 1.10.0
    112     */
    113    private static String algorithmC(final CharSequence left, final CharSequence right) {
    114        final int m = left.length();
    115        final int n = right.length();
    116
    117        final StringBuilder out = new StringBuilder();
    118
    119        if (m == 1) { // Handle trivial cases, as per the paper
    120            final char leftCh = left.charAt(0);
    121            for (int j = 0; j < n; j++) {
    122                if (leftCh == right.charAt(j)) {
    123                    out.append(leftCh);
    124                    break;
    125                }
    126            }
    127        } else if (n > 0 && m > 1) {
    128            final int mid = m / 2; // Find the middle point
    129
    130            final CharSequence leftFirstPart = left.subSequence(0, mid);
    131            final CharSequence leftSecondPart = left.subSequence(mid, m);
    132
    133            // Step 3 of the algorithm: two calls to Algorithm B
    134            final int[] l1 = algorithmB(leftFirstPart, right);
    135            final int[] l2 = algorithmB(reverse(leftSecondPart), reverse(right));
    136
    137            // Find k, as per the Step 4 of the algorithm
    138            int k = 0;
    139            int t = 0;
    140            for (int j = 0; j <= n; j++) {
    141                final int s = l1[j] + l2[n - j];
    142                if (t < s) {
    143                    t = s;
    144                    k = j;
    145                }
    146            }
    147
    148            // Step 5: solve simpler problems, recursively
    149            out.append(algorithmC(leftFirstPart, right.subSequence(0, k)));
    150            out.append(algorithmC(leftSecondPart, right.subSequence(k, n)));
    151        }
    152
    153        return out.toString();
    154    }
    155
    156    // An auxiliary method for CharSequence reversal
    157    private static String reverse(final CharSequence s) {
    158        return new StringBuilder(s).reverse().toString();
    159    }
    160
    161    /**
    162     * Calculates the longest common subsequence similarity score of two {@code CharSequence}'s passed as
    163     * input.
    164     *
    165     * <p>
    166     * This method implements a more efficient version of LCS algorithm which has quadratic time and
    167     * linear space complexity.
    168     * </p>
    169     *
    170     * <p>
    171     * This method is based on newly implemented {@link #algorithmB(CharSequence, CharSequence)}.
    172     * An evaluation using JMH revealed that this method is almost two times faster than its previous version.
    173     * </p>
    174     *
    175     * @param left first character sequence
    176     * @param right second character sequence
    177     * @return length of the longest common subsequence of <code>left</code> and <code>right</code>
    178     * @throws IllegalArgumentException if either String input {@code null}
    179     */
    180    @Override
    181    public Integer apply(final CharSequence left, final CharSequence right) {
    182        // Quick return for invalid inputs
    183        if (left == null || right == null) {
    184            throw new IllegalArgumentException("Inputs must not be null");
    185        }
    186        // Find lengths of two strings
    187        final int leftSz = left.length();
    188        final int rightSz = right.length();
    189
    190        // Check if we can avoid calling algorithmB which involves heap space allocation
    191        if (leftSz == 0 || rightSz == 0) {
    192            return 0;
    193        }
    194
    195        // Check if we can save even more space
    196        if (leftSz < rightSz) {
    197            return algorithmB(right, left)[leftSz];
    198        }
    199        return algorithmB(left, right)[rightSz];
    200    }
    201
    202    /**
    203     * Computes the longest common subsequence between the two {@code CharSequence}'s passed as input.
    204     *
    205     * <p>
    206     * Note, a substring and subsequence are not necessarily the same thing. Indeed, {@code abcxyzqrs} and
    207     * {@code xyzghfm} have both the same common substring and subsequence, namely {@code xyz}. However,
    208     * {@code axbyczqrs} and {@code abcxyzqtv} have the longest common subsequence {@code xyzq} because a
    209     * subsequence need not have adjacent characters.
    210     * </p>
    211     *
    212     * <p>
    213     * For reference, we give the definition of a subsequence for the reader: a <i>subsequence</i> is a sequence that
    214     * can be derived from another sequence by deleting some elements without changing the order of the remaining
    215     * elements.
    216     * </p>
    217     *
    218     * @param left first character sequence
    219     * @param right second character sequence
    220     * @return the longest common subsequence found
    221     * @throws IllegalArgumentException if either String input {@code null}
    222     * @deprecated Deprecated as of 1.2 due to a typo in the method name.
    223     * Use {@link #longestCommonSubsequence(CharSequence, CharSequence)} instead.
    224     * This method will be removed in 2.0.
    225     */
    226    @Deprecated
    227    public CharSequence logestCommonSubsequence(final CharSequence left, final CharSequence right) {
    228        return longestCommonSubsequence(left, right);
    229    }
    230
    231    /**
    232     * Computes the longest common subsequence between the two {@code CharSequence}'s passed as
    233     * input.
    234     *
    235     * <p>
    236     * This method implements a more efficient version of LCS algorithm which although has quadratic time, it
    237     * has linear space complexity.
    238     * </p>
    239     *
    240     *
    241     * <p>
    242     * Note, a substring and subsequence are not necessarily the same thing. Indeed, {@code abcxyzqrs} and
    243     * {@code xyzghfm} have both the same common substring and subsequence, namely {@code xyz}. However,
    244     * {@code axbyczqrs} and {@code abcxyzqtv} have the longest common subsequence {@code xyzq} because a
    245     * subsequence need not have adjacent characters.
    246     * </p>
    247     *
    248     * <p>
    249     * For reference, we give the definition of a subsequence for the reader: a <i>subsequence</i> is a sequence that
    250     * can be derived from another sequence by deleting some elements without changing the order of the remaining
    251     * elements.
    252     * </p>
    253     *
    254     * @param left first character sequence
    255     * @param right second character sequence
    256     * @return the longest common subsequence found
    257     * @throws IllegalArgumentException if either String input {@code null}
    258     * @since 1.2
    259     */
    260    public CharSequence longestCommonSubsequence(final CharSequence left, final CharSequence right) {
    261        // Quick return
    262        if (left == null || right == null) {
    263            throw new IllegalArgumentException("Inputs must not be null");
    264        }
    265        // Find lengths of two strings
    266        final int leftSz = left.length();
    267        final int rightSz = right.length();
    268
    269        // Check if we can avoid calling algorithmC which involves heap space allocation
    270        if (leftSz == 0 || rightSz == 0) {
    271            return "";
    272        }
    273
    274        // Check if we can save even more space
    275        if (leftSz < rightSz) {
    276            return algorithmC(right, left);
    277        }
    278        return algorithmC(left, right);
    279    }
    280
    281    /**
    282     * Computes the lcsLengthArray for the sake of doing the actual lcs calculation. This is the
    283     * dynamic programming portion of the algorithm, and is the reason for the runtime complexity being
    284     * O(m*n), where m=left.length() and n=right.length().
    285     *
    286     * @param left first character sequence
    287     * @param right second character sequence
    288     * @return lcsLengthArray
    289     * @deprecated Deprecated as of 1.10. A more efficient implementation for calculating LCS is now available.
    290     * Use {@link #longestCommonSubsequence(CharSequence, CharSequence)} instead to directly calculate the LCS.
    291     * This method will be removed in 2.0.
    292     */
    293    @Deprecated
    294    public int[][] longestCommonSubstringLengthArray(final CharSequence left, final CharSequence right) {
    295        final int[][] lcsLengthArray = new int[left.length() + 1][right.length() + 1];
    296        for (int i = 0; i < left.length(); i++) {
    297            for (int j = 0; j < right.length(); j++) {
    298                if (i == 0) {
    299                    lcsLengthArray[i][j] = 0;
    300                }
    301                if (j == 0) {
    302                    lcsLengthArray[i][j] = 0;
    303                }
    304                if (left.charAt(i) == right.charAt(j)) {
    305                    lcsLengthArray[i + 1][j + 1] = lcsLengthArray[i][j] + 1;
    306                } else {
    307                    lcsLengthArray[i + 1][j + 1] = Math.max(lcsLengthArray[i + 1][j], lcsLengthArray[i][j + 1]);
    308                }
    309            }
    310        }
    311        return lcsLengthArray;
    312    }
    313
    314}