[leet code] Edit Distance
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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
=====================
Analysis:
DP problem!
Base case:
str1.length == 0 ; str2.length == 0; => operation=0;
str1.length == k ; str2.length == 0; => operation=k; (delete characters from str1)
str1.length == 0; str2.length == k; => operation=k; (insert characters to str1)
After we have the base case, when we are calculating Edit Distance(str1+char1, str2+char2), we will have cases below:
if (char1 == char2) Edit Distance(str1+char1, str2+char2) = Edit Distance(str1, str2) (no additional operation is needed)
if (char1 != char2){
3 ways to calculate Edit Distance(str1+char1, str2+char2):
Edit Distance(str1+char1, str2+char2) = Edit Distance(str1, str2) + 1 (+1 here means to replace char1 by char2)
Edit Distance(str1+char1, str2+char2) = Edit Distance(str1+char1, str2) + 1 (+1 here means to insert char2 to str2)
Edit Distance(str1+char1, str2+char2) = Edit Distance(str1, str2+char2) + 1 (+1 here means to delete char1 from str1)
}
public class Solution { public int minDistance(String word1, String word2) { int[][] dp = new int[word1.length()+1][word2.length()+1]; for(int i=0; i<dp.length; i++) dp[i][0] = i; // insert to word1 operations only for(int i=0; i<dp[0].length; i++) dp[0][i] = i; // delete from word1 operations only for(int i=1; i<=word1.length(); i++){ for(int j=1; j<=word2.length(); j++){ if(word1.charAt(i-1)==word2.charAt(j-1)) dp[i][j] = dp[i-1][j-1]; else{ int replace=1+dp[i-1][j-1]; int insert=1+dp[i][j-1]; int delete=1+dp[i-1][j]; dp[i][j]=Math.min((Math.min(replace, insert)), delete); } } } return dp[word1.length()][word2.length()]; }}
Ref:
http://www.youtube.com/watch?v=CB425OsE4Fo&list=PL0174E49C0E0DD5C8&index=35
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