[LeetCode] 72. 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

Thought process:
  1. Sub-problem: find the edit distance between word1[0, i] and word2[0, j].
  2. Function: 
    1. If word1[i] == word2[j], we can be greedy and f[i][j] = f[i - 1][j - 1].
    2. Else, f[i][j] = min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1
  3. Initialization: f[i][0] = i and f[0][i] = i because any word's edit distance with an empty word is the length of the word.
  4. Answer: the answer is at f[word1.length][word2.length].

Solution:
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public class Solution {
    public int minDistance(String word1, String word2) {
        int length1 = word1.length();
        int length2 = word2.length();
        
        int[][] f = new int[length1 + 1][length2 + 1];
        
        for (int i = 0; i <= length1; i++) {
            f[i][0] = i;
        }
        for (int i = 1; i <= length2; i++) {
            f[0][i] = i;
        }
        
        for (int i = 1; i <= length1; i++) {
            for (int j = 1; j <= length2; j++) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    f[i][j] = f[i - 1][j - 1];
                } else {
                    f[i][j] = Math.min(f[i - 1][j - 1], Math.min(f[i - 1][j], f[i][j - 1])) + 1;
                }
            }
        }
        
        return f[length1][length2];
    }
}

Time complexity:
Say the lengths of word1 and word2 are a and b. The time complexity is O(ab).

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