Hard
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:
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.class Solution {
int minDistance(String w1, String w2) {
int n1 = w1.length;
int n2 = w2.length;
// If w2 is longer, swap w1 and w2 to ensure w1 is the longer string
if (n2 > n1) {
return minDistance(w2, w1);
}
List<int> dp = List.filled(n2 + 1, 0);
// Initialize the first row of the DP table
for (int j = 0; j <= n2; j++) {
dp[j] = j;
}
for (int i = 1; i <= n1; i++) {
int pre = dp[0];
dp[0] = i;
for (int j = 1; j <= n2; j++) {
int temp = dp[j];
dp[j] = w1[i - 1] != w2[j - 1]
? 1 + [pre, dp[j], dp[j - 1]].reduce((a, b) => a < b ? a : b)
: pre;
pre = temp;
}
}
return dp[n2];
}
}