72. Edit Distance
1. Description
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
2. Example
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’)
3. Constraints
- 0 <= word1.length, word2.length <= 500
- word1 and word2 consist of lowercase English letters.
4. Solutions
Dynamic Programming
m = word1.size(), n = word2.size()
Time complexity: O(mn)
Space complexity: O(mn)
class Solution {
public:
int minDistance(const string &word1, const string &word2) {
if (word1.size() * word2.size() == 0) {
return word1.size() + word2.size();
}
vector<vector<int>> min_dist(word1.size() + 1, vector<int>(word2.size() + 1));
for (int i = 0; i <= word1.size(); ++i) {
min_dist[i][0] = i;
}
for (int i = 0; i <= word2.size(); ++i) {
min_dist[0][i] = i;
}
for (int i = 1; i <= word1.size(); ++i) {
for (int j = 1; j <= word2.size(); ++j) {
if (word1[i - 1] == word2[j - 1]) {
min_dist[i][j] = min(
1 + min(min_dist[i - 1][j], min_dist[i][j - 1]), min_dist[i - 1][j - 1]);
} else {
min_dist[i][j] =
1 + min({min_dist[i - 1][j], min_dist[i][j - 1], min_dist[i - 1][j - 1]});
}
}
}
return min_dist.back().back();
}
};