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(min(m, n))
class Solution {
public:
int minDistance(const string &word1, const string &word2) {
bool word1_longer = word1.size() > word2.size();
const string w1 = word1_longer ? word1 : word2;
const string w2 = word1_longer ? word2 : word1;
const int m = w1.size(), n = w2.size();
vector<int> count(n + 1, 0);
for (int i = 0; i <= n; ++i) {
count[i] = i;
}
for (int i = 1; i <= m; ++i) {
int diagonal = count[0];
count[0] = i;
for (int j = 1; j <= n; ++j) {
int above = count[j];
if (w1[i - 1] == w2[j - 1]) {
count[j] = diagonal;
} else {
count[j] = 1 + min(diagonal, min(count[j], count[j - 1]));
}
diagonal = above;
}
}
return count.back();
}
};