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EditDistance.java
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EditDistance.java
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package Leetcode;
/**
* @author kalpak
*
* Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.
*
* You have the following three 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')
*
*
* Constraints:
*
* 0 <= word1.length, word2.length <= 500
* word1 and word2 consist of lowercase English letters.
*
*/
public class EditDistance {
public static int minDistance(String word1, String word2) {
/**
* If the characters are equal, no need for any operation. So copy the diagonal value which is dp[i - 1][j - 1].
* Else, Check for the minimum amongst. top, diagonal left, and left - and add 1 to it.
*/
int[][] dp = new int[word1.length() + 1][word2.length() + 2];
for(int i = 0; i < dp.length; i++)
dp[i][0] = i;
for(int i = 0; i < dp[0].length; i++)
dp[0][i] = i;
for(int i = 1; i < word1.length() + 1; i++) {
for(int j = 1; j < word2.length() + 1; j++) {
if(word1.charAt(i - 1) == word2.charAt(j - 1))
dp[i][j] = dp[i - 1][j - 1];
else
dp[i][j] = Math.min(dp[i - 1][j], // Remove
Math.min(dp[i - 1][j - 1], // Replace
dp[i][j - 1])) // Insert
+ 1;
}
}
return dp[word1.length()][word2.length()];
}
}