package DynamicProgramming;

/**
 * A DynamicProgramming based solution for Edit Distance problem In Java Description of Edit
 * Distance with an Example:
 *
 * <p>Edit distance is a way of quantifying how dissimilar two strings (e.g., words) are to one
 * another, by counting the minimum number of operations required to transform one string into the
 * other. The distance operations are the removal, insertion, or substitution of a character in the
 * string.
 *
 * <p>
 *
 * <p>The Distance between "kitten" and "sitting" is 3. A minimal edit script that transforms the
 * former into the latter is:
 *
 * <p>kitten → sitten (substitution of "s" for "k") sitten → sittin (substitution of "i" for "e")
 * sittin → sitting (insertion of "g" at the end).
 *
 * @author SUBHAM SANGHAI
 */
import java.util.Scanner;

public class EditDistance {

  public static int minDistance(String word1, String word2) {
    int len1 = word1.length();
    int len2 = word2.length();
    // len1+1, len2+1, because finally return dp[len1][len2]
    int[][] dp = new int[len1 + 1][len2 + 1];
    /* If second string is empty, the only option is to
    insert all characters of first string into second*/
    for (int i = 0; i <= len1; i++) {
      dp[i][0] = i;
    }
    /* If first string is empty, the only option is to
    insert all characters of second string into first*/
    for (int j = 0; j <= len2; j++) {
      dp[0][j] = j;
    }
    // iterate though, and check last char
    for (int i = 0; i < len1; i++) {
      char c1 = word1.charAt(i);
      for (int j = 0; j < len2; j++) {
        char c2 = word2.charAt(j);
        // if last two chars equal
        if (c1 == c2) {
          // update dp value for +1 length
          dp[i + 1][j + 1] = dp[i][j];
        } else {
          /* if two characters are different ,
          then take the minimum of the various operations(i.e insertion,removal,substitution)*/
          int replace = dp[i][j] + 1;
          int insert = dp[i][j + 1] + 1;
          int delete = dp[i + 1][j] + 1;

          int min = replace > insert ? insert : replace;
          min = delete > min ? min : delete;
          dp[i + 1][j + 1] = min;
        }
      }
    }
    /* return the final answer , after traversing through both the strings*/
    return dp[len1][len2];
  }

  public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    String s1, s2;
    System.out.println("Enter the First String");
    s1 = input.nextLine();
    System.out.println("Enter the Second String");
    s2 = input.nextLine();
    // ans stores the final Edit Distance between the two strings
    int ans = minDistance(s1, s2);
    System.out.println(
        "The minimum Edit Distance between \"" + s1 + "\" and \"" + s2 + "\" is " + ans);
    input.close();
  }

   // edit distance problem

  public static int editDistance(String s1, String s2){
  int[][] storage = new int[s1.length() + 1][s2.length() + 1];
      return editDistance(s1, s2,storage);

  }

  public static int editDistance(String s1, String s2, int[][] storage) {
      int m = s1.length();
      int n = s2.length();
      if (storage[m][n] > 0) {
          return storage[m][n];

}
if (m== 0) {
          storage[m][n] = n;
          return storage[m][n];

      }
      if (n== 0) {
          storage[m][n] = m;
          return storage[m][n];

      }
      if (s1.charAt(0) == s2.charAt(0)) {
          storage[m][n] = editDistance(s1.substring(1), s2.substring(1), storage);
          return storage[m][n];

      } else {
          int op1 = editDistance(s1, s2.substring(1),storage);
          int op2 = editDistance(s1.substring(1), s2,storage);
          int op3 = editDistance(s1.substring(1),s2.substring(1),storage);
          storage[m][n] = 1 + Math.min(op1, Math.min(op2, op3));
          return storage[m][n];
      }
  }
}