72. Edit Distance (Hard)

https://leetcode.com/problems/edit-distance/

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')

Solutions

class Solution {

    public int minDistance(String word1, String word2) {
        if (word1 == null && word2 == null) {
            return 0;
        }

        if (word1 == null) {
            return word2.length();
        }

        if (word2 == null) {
            return word1.length();
        }

        int len1 = word1.length();
        int len2 = word2.length();

        int[][] dp = new int[len1 + 1][len2 + 1];

        // length of word1 is i, word2 is 0, update the edit distance
        for (int i = 0; i <= len1; i++) {
            dp[i][0] = i;
        }

        // length of word2 is i, word1 is 0, update the edit distance
        for (int j = 0; j <= len2; j++) {
            dp[0][j] = j;
        }

        for (int i = 1; i <= len1; i++) {

            // Assume word1's length is i, work out the edit distance when j is 1,2,3,...len2 respectively.
            for (int j = 1; j <= len2; j++) {

                // Following operations can be considered as calculating edit distance from word1[0:i-1] to
                // word2[0:j-1]

                // what if we delete the last char of word1[0:i-1], then we adopt the edit distance from
                // word1[0:i-2] to word2[j-1]
                int del = dp[i - 1][j] + 1;

                // what if we delete the last char of word2[0:j-1], then we adopt the edit distance from
                // word1[0:i-1] to word2[j-2]
                int ins = dp[i][j - 1] + 1;

                // what if we replace word1[i-1] with word2[j-1] if not the same, then we adopt the edit distance
                // from word1[0:i-2] to word2[j-2]
                int rep = dp[i - 1][j - 1];

                // No need to change if the last char is the same, following i-1 and j-1 is the index, the
                // actual meaning is the ith char in word1 and jth char in word2.
                if (word1.charAt(i - 1) != word2.charAt(j - 1)) {
                    rep += 1;
                }

                dp[i][j] = min(del, ins, rep);
            }
        }

        return dp[len1][len2];
    }

    private int min(int x, int y, int z) {
        int min = Math.min(x, y);

        return Math.min(min, z);
    }
}

Incorrect Solutions

References

  1. https://leetcode.com/problems/edit-distance/discuss/25870/My-none-recursive-solution
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