public static int getLevenshteinDistance2 (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];
   }