72. Edit Distance
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.
From: LeetCode
Link: 72. Edit Distance
Solution:
Ideas：
- Initialize base cases of empty strings
- Use dynamic programming to fill table
- Minimum operations at any point is 1 (for replace) + minimum of insert, delete or replace from previous state
Code:
int min(int a, int b){
return a < b ? a : b;
}
int minDistance(char* word1, char* word2) {
int m = strlen(word1);
int n = strlen(word2);
int dp[m+1][n+1];
for(int i = 0; i <= m; i++) {
dp[i][0] = i;
}
for(int j = 0; j <= n; j++) {
dp[0][j] = j;
}
for(int i = 1; i <= m; i++) {
for(int j = 1; j <= n; j++) {
if(word1[i-1] == word2[j-1]) {
dp[i][j] = dp[i-1][j-1];
} else {
dp[i][j] = 1 + min(dp[i][j-1], min(dp[i-1][j], dp[i-1][j-1]));
}
}
}
return dp[m][n];
}
