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];
    }