/*
 * 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.
 */

using System;
using System.Collections.Generic;
using System.Text;
using SpellChecker.Net.Search.Spell;

namespace SpellChecker.Net.Search.Spell
{
    /// <summary>
    /// Levenshtein edit distance
    /// </summary>
    public class LevenshteinDistance : StringDistance
    {
        /// <summary>
        /// Returns a float between 0 and 1 based on how similar the specified strings are to one another.  
        /// Returning a value of 1 means the specified strings are identical and 0 means the
        /// string are maximally different.
        /// </summary>
        /// <param name="target">The first string.</param>
        /// <param name="other">The second string.</param>
        /// <returns>a float between 0 and 1 based on how similar the specified strings are to one another.</returns>
        public float GetDistance(String target, String other)
        {
            char[] sa;
            int n;
            int[] p; //'previous' cost array, horizontally
            int[] d; // cost array, horizontally
            int[] _d; //placeholder to assist in swapping p and d

            /*
               The difference between this impl. and the previous is that, rather
               than creating and retaining a matrix of size s.length()+1 by t.length()+1,
               we maintain two single-dimensional arrays of length s.length()+1.  The first, d,
               is the 'current working' distance array that maintains the newest distance cost
               counts as we iterate through the characters of String s.  Each time we increment
               the index of String t we are comparing, d is copied to p, the second int[].  Doing so
               allows us to retain the previous cost counts as required by the algorithm (taking
               the minimum of the cost count to the left, up one, and diagonally up and to the left
               of the current cost count being calculated).  (Note that the arrays aren't really
               copied anymore, just switched...this is clearly much better than cloning an array
               or doing a System.arraycopy() each time  through the outer loop.)

               Effectively, the difference between the two implementations is this one does not
               cause an out of memory condition when calculating the LD over two very large strings.
             */

            sa = target.ToCharArray();
            n = sa.Length;
            p = new int[n + 1];
            d = new int[n + 1];
            int m = other.Length;

            if (n == 0 || m == 0)
            {
                if (n == m)
                {
                    return 1;
                }
                else
                {
                    return 0;
                }
            }


            // indexes into strings s and t
            int i; // iterates through s
            int j; // iterates through t

            char t_j; // jth character of t

            int cost; // cost

            for (i = 0; i <= n; i++)
            {
                p[i] = i;
            }

            for (j = 1; j <= m; j++)
            {
                t_j = other[j - 1];
                d[0] = j;

                for (i = 1; i <= n; i++)
                {
                    cost = sa[i - 1] == t_j ? 0 : 1;
                    // minimum of cell to the left+1, to the top+1, diagonally left and up +cost
                    d[i] = Math.Min(Math.Min(d[i - 1] + 1, p[i] + 1), p[i - 1] + cost);
                }

                // copy current distance counts to 'previous row' distance counts
                _d = p;
                p = d;
                d = _d;
            }

            // our last action in the above loop was to switch d and p, so p now
            // actually has the most recent cost counts
            return 1.0f - ((float)p[n] / Math.Max(other.Length, sa.Length));
        }
    }
}
