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