/**
 * Copyright 2008 - CommonCrawl Foundation
 * 
 *    This program is free software: you can redistribute it and/or modify
 *    it under the terms of the GNU General Public License as published by
 *    the Free Software Foundation, either version 3 of the License, or
 *    (at your option) any later version.
 *
 *    This program is distributed in the hope that it will be useful,
 *    but WITHOUT ANY WARRANTY; without even the implied warranty of
 *    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 *    GNU General Public License for more details.
 *
 *    You should have received a copy of the GNU General Public License
 *    along with this program.  If not, see <http://www.gnu.org/licenses/>.
 *
 **/

package org.commoncrawl.util;

/**
 * 
 * @author rana
 *
 */
public class StringUtils {
  
  public static int getLevenshteinDistance (String s, String t) {
    if (s == null || t == null) {
      throw new IllegalArgumentException("Strings must not be null");
    }
          
    /*
      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.          
    */        
          
    int n = s.length(); // length of s
    int m = t.length(); // length of t
          
    if (n == 0) {
      return m;
    } else if (m == 0) {
      return n;
    }

    int p[] = new int[n+1]; //'previous' cost array, horizontally
    int d[] = new int[n+1]; // cost array, horizontally
    int _d[]; //placeholder to assist in swapping p and d

    // 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 = t.charAt(j-1);
       d[0] = j;
          
       for (i=1; i<=n; i++) {
          cost = s.charAt(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 p[n];
  }
}