Edit Distance

LeetCode Q 72 - 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:

• Insert a character
• Delete a character
• 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')

Solution : DP

dp[i][j]: the distance between ith char in word1 and jth char in word2

• if (word1.charAt(i) == word1.charAt(j)) dp[i][j] = dp[i - 1][j - 1];
• else we can get to j from i though
  • replace a char, dp[i][j] = dp[i - 1][j - 1]
  • delete a char, dp[i][j] = dp[i][j - 1];
  • insert a char, dp[i][j] = dp[i - 1][j];

public int minDistance(String word1, String word2) {
	if (word1.length() == 0) return word2.length(); // insert
	if (word2.length() == 0) return word1.length(); // delete
	
	int len1 = word1.length(), len2 = word2.length();
	// dp[i][j]: the distance between ith char in word1 and jth char in word2
	int[][] dp = new int[len1 + 1][len2 + 1];
	// word1 is ""
	for (int i = 0; i <= len2; i++) 
		dp[0][i] = i; 
	// word2 is ""
	for (int i = 0; i <= len1; i++) 
		dp[i][0] = i;
	
	for (int i = 1; i <= len1; i++) {
		for (int j = 1; j <= len2; j++) {
			if (word1.charAt(i - 1) == word2.charAt(j - 1))
				dp[i][j] = dp[i - 1][j - 1];
			else {
				dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1], dp[i][j - 1]), dp[i - 1][j]) + 1;
			}
		}
	}
	return dp[len1][len2];
}