LeetCode - Edit Distance

Problem description

description

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

1. Insert a character
2. Delete a character
3. Replace a character Example 1:

   Input: word1 = "horse", word2 = "ros"
   Output: 3
   Explanation: 
   horse -> rorse (replace 'h' with 'r')
   rorse -> rose (remove 'r')
   rose -> ros (remove 'e')
   

   Example 2:

   Input: word1 = "intention", word2 = "execution"
   Output: 5
   Explanation: 
   intention -> inention (remove 't')
   inention -> enention (replace 'i' with 'e')
   enention -> exention (replace 'n' with 'x')
   exention -> exection (replace 'n' with 'c')
   exection -> execution (insert 'u')
   

Analysis

The first thought is to use backtrack with wildcard. But the correct solution is to use DP.

class Solution {
    public int minDistance(String word1, String word2) {
        int len1 = word1.length();
        int len2 = word2.length();
        
        if (len1 == 0 || len2 == 0){
            return len1+len2;
        }
        
        int[][] dp = new int[len1][len2];
        
        dp[0][0] = word1.charAt(0) == word2.charAt(0) ? 0:1;
        
        for (int i = 0; i < len1; i++){
            for (int j = 0; j < len2; j++){
                if (i == 0 && j == 0){
                    continue;
                }
                
                boolean equal = word1.charAt(i) == word2.charAt(j);
                
                if (i == 0){
                    dp[i][j] = equal ? j : dp [i][j-1] + 1; // should be j
                }
                else if (j == 0){
                    dp[i][j] = equal ? i : dp[i-1][j] + 1; // should be i
                }
                else {
                    dp[i][j] = equal ? dp[i-1][j-1] : Math.min(dp[i][j-1],Math.min(dp[i-1][j], dp[i-1][j-1])) + 1;
                }
            }
        }
        
        return dp[len1-1][len2-1];
    }
}

What to improve

• ask for count but not details we should use dp

• be careful about the base case.