# [leet code] Edit Distance

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

=====================

Analysis:

DP problem!

Base case:

str1.length == 0 ; str2.length == 0; => operation=0;

str1.length == k ; str2.length == 0; => operation=k; (delete characters from str1)

str1.length == 0; str2.length == k; => operation=k; (insert characters to str1)

After we have the base case, when we are calculating Edit Distance(str1+char1, str2+char2), we will have cases below:

if (char1 == char2) Edit Distance(str1+char1, str2+char2) = Edit Distance(str1, str2) (no additional operation is needed)

if (char1 != char2){

3 ways to calculate Edit Distance(str1+char1, str2+char2):

Edit Distance(str1+char1, str2+char2) = Edit Distance(str1, str2) + 1 (+1 here means to replace char1 by char2)

Edit Distance(str1+char1, str2+char2) = Edit Distance(str1+char1, str2) + 1 (+1 here means to insert char2 to str2)

Edit Distance(str1+char1, str2+char2) = Edit Distance(str1, str2+char2) + 1 (+1 here means to delete char1 from str1)

}

`public class Solution {    public int minDistance(String word1, String word2) {        int[][] dp = new int[word1.length()+1][word2.length()+1];                for(int i=0; i<dp.length; i++) dp[i][0] = i; // insert to word1 operations only        for(int i=0; i<dp[0].length; i++) dp[0][i] = i; // delete from word1 operations only                for(int i=1; i<=word1.length(); i++){            for(int j=1; j<=word2.length(); j++){                if(word1.charAt(i-1)==word2.charAt(j-1)) dp[i][j] = dp[i-1][j-1];                else{                    int replace=1+dp[i-1][j-1];                    int insert=1+dp[i][j-1];                    int delete=1+dp[i-1][j];                    dp[i][j]=Math.min((Math.min(replace, insert)), delete);                }            }        }                return dp[word1.length()][word2.length()];    }}`

Ref: