Edit Distance (Levenshtein)

/*
 * Problem : Find the edit distance.
 * Input   : Two words.
 * Output  : The minimum number of operations.
 * Approach: Compare prefixes.
 * Time    : O(m * n)
 * Space   : O(n)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static int solve(String a, String b) {
        // Step 1: previous[j] = the edit distance of the first i - 1 letters of a and the first j letters of b.
        // Changing an empty word to the first j letters of b needs j insertions.
        int[] previous = new int[b.length() + 1];
        for (int j = 0; j <= b.length(); j++) {
            previous[j] = j;
        }

        for (int i = 1; i <= a.length(); i++) {
            // Step 2: current[0] = i, because we must delete i letters to get an empty word.
            int[] current = new int[b.length() + 1];
            current[0] = i;

            for (int j = 1; j <= b.length(); j++) {
                if (a.charAt(i - 1) == b.charAt(j - 1)) {
                    // Step 3: Equal letters cost nothing.
                    current[j] = previous[j - 1];
                } else {
                    // Step 4: Otherwise take the cheapest of: replace, delete, insert. Each one costs 1 more.
                    current[j] = 1 + Math.min(previous[j - 1], Math.min(previous[j], current[j - 1]));
                }
            }
            previous = current;
        }
        return previous[b.length()];
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String a = scanner.next();
        String b = scanner.next();
        System.out.println(solve(a, b));
    }
}