/*
    * Licensed to the Apache Software Foundation (ASF) under one or more
    * contributor license agreements.  See the NOTICE file distributed with
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    * 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.
   *
   * <p>
   * 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.
   * <i>Note.</i>  Generally this algorithm is fairly inefficient, as for length <i>m</i>, <i>n</i> of the input
   * {@code CharSequence}'s {@code left} and {@code right} respectively, the runtime of the
   * algorithm is <i>O(m*n)</i>.
   * </p>
   *
   * <p>
   * As of version 1.10, a more space-efficient of the algorithm is implemented. The new algorithm has linear space
   * complexity instead of quadratic. However, time complexity is still quadratic in the size of input strings.
   * </p>
   *
   * <p>
   * The implementation is based on Hirschberg's Longest Commons Substring algorithm (cited below).
   * </p>
   *
   * <p>For further reading see:</p>
   * <ul>
   * <li>
   * Lothaire, M. <i>Applied combinatorics on words</i>. New York: Cambridge U Press, 2005. <b>12-13</b>
   * </li>
   * <li>
   * D. S. Hirschberg, "A linear space algorithm for computing maximal common subsequences," CACM, 1975, pp. 341--343.
   * </li>
   * </ul>
   *
   * @since 1.0
   */
  public class LongestCommonSubsequence implements SimilarityScore<Integer> {
  
      /**
       * Singleton instance.
       */
      static final LongestCommonSubsequence INSTANCE = new LongestCommonSubsequence();
  
      /**
       * An implementation of "ALG B" from Hirschberg's CACM '71 paper.
       * Assuming the first input sequence is of size <code>m</code> and the second input sequence is of size
       * <code>n</code>, this method returns the last row of the dynamic programming (DP) table when calculating
       * the LCS of the two sequences in <i>O(m*n)</i> time and <i>O(n)</i> space.
       * The last element of the returned array, is the size of the LCS of the two input sequences.
       *
       * @param left first input sequence.
       * @param right second input sequence.
       * @return last row of the dynamic-programming (DP) table for calculating the LCS of <code>left</code> and <code>right</code>
       * @since 1.10.0
       */
      private static int[] algorithmB(final CharSequence left, final CharSequence right) {
          final int m = left.length();
          final int n = right.length();
  
          // Creating an array for storing two rows of DP table
          final int[][] dpRows = new int[2][1 + n];
  
          for (int i = 1; i <= m; i++) {
              // K(0, j) <- K(1, j) [j = 0...n], as per the paper:
              // Since we have references in Java, we can swap references instead of literal copying.
              // We could also use a "binary index" using modulus operator, but directly swapping the
              // two rows helps readability and keeps the code consistent with the algorithm description
              // in the paper.
              final int[] temp = dpRows[0];
              dpRows[0] = dpRows[1];
              dpRows[1] = temp;
  
              for (int j = 1; j <= n; j++) {
                  if (left.charAt(i - 1) == right.charAt(j - 1)) {
                      dpRows[1][j] = dpRows[0][j - 1] + 1;
                  } else {
                      dpRows[1][j] = Math.max(dpRows[1][j - 1], dpRows[0][j]);
                  }
              }
          }
  
          // LL(j) <- K(1, j) [j=0...n], as per the paper:
          // We don't need literal copying of the array, we can just return the reference
          return dpRows[1];
     }
 
     /**
      * An implementation of "ALG C" from Hirschberg's CACM '71 paper.
      * Assuming the first input sequence is of size <code>m</code> and the second input sequence is of size
      * <code>n</code>, this method returns the Longest Common Subsequence (LCS) of the two sequences in
      * <i>O(m*n)</i> time and <i>O(m+n)</i> space.
      *
      * @param left first input sequence.
      * @param right second input sequence.
      * @return the LCS of <code>left</code> and <code>right</code>
      * @since 1.10.0
      */
     private static String algorithmC(final CharSequence left, final CharSequence right) {
         final int m = left.length();
         final int n = right.length();
 
         final StringBuilder out = new StringBuilder();
 
         if (m == 1) { // Handle trivial cases, as per the paper
             final char leftCh = left.charAt(0);
             for (int j = 0; j < n; j++) {
                 if (leftCh == right.charAt(j)) {
                     out.append(leftCh);
                     break;
                 }
             }
         } else if (n > 0 && m > 1) {
             final int mid = m / 2; // Find the middle point
 
             final CharSequence leftFirstPart = left.subSequence(0, mid);
             final CharSequence leftSecondPart = left.subSequence(mid, m);
 
             // Step 3 of the algorithm: two calls to Algorithm B
             final int[] l1 = algorithmB(leftFirstPart, right);
             final int[] l2 = algorithmB(reverse(leftSecondPart), reverse(right));
 
             // Find k, as per the Step 4 of the algorithm
             int k = 0;
             int t = 0;
             for (int j = 0; j <= n; j++) {
                 final int s = l1[j] + l2[n - j];
                 if (t < s) {
                     t = s;
                     k = j;
                 }
             }
 
             // Step 5: solve simpler problems, recursively
             out.append(algorithmC(leftFirstPart, right.subSequence(0, k)));
             out.append(algorithmC(leftSecondPart, right.subSequence(k, n)));
         }
 
         return out.toString();
     }
 
     // An auxiliary method for CharSequence reversal
     private static String reverse(final CharSequence s) {
         return new StringBuilder(s).reverse().toString();
     }
 
     /**
      * Calculates the longest common subsequence similarity score of two {@code CharSequence}'s passed as
      * input.
      *
      * <p>
      * This method implements a more efficient version of LCS algorithm which has quadratic time and
      * linear space complexity.
      * </p>
      *
      * <p>
      * This method is based on newly implemented {@link #algorithmB(CharSequence, CharSequence)}.
      * An evaluation using JMH revealed that this method is almost two times faster than its previous version.
      * </p>
      *
      * @param left first character sequence
      * @param right second character sequence
      * @return length of the longest common subsequence of <code>left</code> and <code>right</code>
      * @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");
         }
         // Find lengths of two strings
         final int leftSz = left.length();
         final int rightSz = right.length();
 
         // Check if we can avoid calling algorithmB which involves heap space allocation
         if (leftSz == 0 || rightSz == 0) {
             return 0;
         }
 
         // Check if we can save even more space
         if (leftSz < rightSz) {
             return algorithmB(right, left)[leftSz];
         }
         return algorithmB(left, right)[rightSz];
     }
 
     /**
      * Computes the longest common subsequence between the two {@code CharSequence}'s passed as input.
      *
      * <p>
      * Note, a substring and subsequence are not necessarily the same thing. Indeed, {@code abcxyzqrs} and
      * {@code xyzghfm} have both the same common substring and subsequence, namely {@code xyz}. However,
      * {@code axbyczqrs} and {@code abcxyzqtv} have the longest common subsequence {@code xyzq} because a
      * subsequence need not have adjacent characters.
      * </p>
      *
      * <p>
      * For reference, we give the definition of a subsequence for the reader: a <i>subsequence</i> is a sequence that
      * can be derived from another sequence by deleting some elements without changing the order of the remaining
      * elements.
      * </p>
      *
      * @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 {@code CharSequence}'s passed as
      * input.
      *
      * <p>
      * This method implements a more efficient version of LCS algorithm which although has quadratic time, it
      * has linear space complexity.
      * </p>
      *
      *
      * <p>
      * Note, a substring and subsequence are not necessarily the same thing. Indeed, {@code abcxyzqrs} and
      * {@code xyzghfm} have both the same common substring and subsequence, namely {@code xyz}. However,
      * {@code axbyczqrs} and {@code abcxyzqtv} have the longest common subsequence {@code xyzq} because a
      * subsequence need not have adjacent characters.
      * </p>
      *
      * <p>
      * For reference, we give the definition of a subsequence for the reader: a <i>subsequence</i> is a sequence that
      * can be derived from another sequence by deleting some elements without changing the order of the remaining
      * elements.
      * </p>
      *
      * @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");
         }
         // Find lengths of two strings
         final int leftSz = left.length();
         final int rightSz = right.length();
 
         // Check if we can avoid calling algorithmC which involves heap space allocation
         if (leftSz == 0 || rightSz == 0) {
             return "";
         }
 
         // Check if we can save even more space
         if (leftSz < rightSz) {
             return algorithmC(right, left);
         }
         return algorithmC(left, right);
     }
 
     /**
      * 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
      * @deprecated Deprecated as of 1.10. A more efficient implementation for calculating LCS is now available.
      * Use {@link #longestCommonSubsequence(CharSequence, CharSequence)} instead to directly calculate the LCS.
      * This method will be removed in 2.0.
      */
     @Deprecated
     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;
     }
 
 }