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 = long_word.size(), n = short_word.size()
Time complexity: O(mn)
Space complexity: O(n)
class Solution {
public:
int minDistance(const string &word1, const string &word2) {
const bool word1_is_longer = word1.size() > word2.size();
const string &long_word = word1_is_longer ? word1 : word2;
const string &short_word = word1_is_longer ? word2 : word1;
const int m = long_word.size(), n = short_word.size();
if(n == 0) {
return m;
}
vector<int> min_distance(n + 1);
for (int j = 0; j <= n; ++j) {
min_distance[j] = j;
}
for (int i = 0; i < m; ++i) {
min_distance[0] = i;
int prev = i;
for (int j = 0; j < n; ++j) {
int temp = min_distance[j + 1];
if (long_word[i] == short_word[j]) {
min_distance[j + 1] = min({prev, 1 + min_distance[j], 1 + min_distance[j + 1]});
} else {
min_distance[j + 1] = 1 + min({prev, min_distance[j], min_distance[j + 1]});
}
prev = temp;
}
}
return min_distance.back();
}
};