Amazon
Bloomberg
Square
Uber
2020-05-31
72. Edit Distance
Question:
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:
Using DP to solve the questions. If character i and j are same, using dp[i-1][j-1]
. Otherwise, there will be three cases:
- Insert -
dp[2][2] = dp[2][1] + 1
For example: edit(“ab”, “ac”) = 1 + edit(“ab”, “a”) <- insert “c” at the end of word1 - Delete -
dp[2][2] = dp[1][2] + 1
For example: edit(“ab”, “ac”) = 1 + edit(“a”, “ac”) <- delete “b” at the end of word1 - Replace -
dp[2][2] = dp[1][1] + 1
For example: edit(“ab”, “ac”) = 1 + edit(“a”, “a”) <- replace “b” by “c” at the end of word1
The following picture demo borrowed from Huahua.
class Solution {
public int minDistance(String word1, String word2) {
int n = word1.length();
int m = word2.length();
int[][] dp = new int[n+1][m+1];
for (int i = 0; i <= n; i++) {
dp[i][0] = i;
}
for (int i = 0; i <= m; i++) {
dp[0][i] = i;
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[i][j] = dp[i-1][j-1];
} else{
dp[i][j] = 1 + Math.min(dp[i-1][j-1], Math.min(dp[i-1][j], dp[i][j-1])); // Replace, Delete, Insert
}
}
}
return dp[n][m];
}
}