72. Edit Distance

2024, Nov 04 2 mins read

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))

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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();
    }
};

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