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Java-based LeetCode algorithm problem solutions, regularly updated
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72\. Edit Distance
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:
* Insert a character
* Delete a character
* Replace a character
**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.
To solve the "Edit Distance" problem in Java with the Solution class, follow these steps:
1. Define a method `minDistance` in the `Solution` class that takes two strings `word1` and `word2` as input and returns the minimum number of operations required to convert `word1` to `word2`.
2. Initialize a 2D array `dp` of size `(m+1) x (n+1)`, where `m` is the length of `word1` and `n` is the length of `word2`.
3. Set `dp[i][0] = i` for all `i` from `0` to `m`, as the minimum number of operations to convert a string of length `i` to an empty string is `i` deletions.
4. Set `dp[0][j] = j` for all `j` from `0` to `n`, as the minimum number of operations to convert an empty string to a string of length `j` is `j` insertions.
5. Iterate over the characters of `word1` and `word2`:
- If `word1.charAt(i-1)` is equal to `word2.charAt(j-1)`, set `dp[i][j] = dp[i-1][j-1]`, as no operation is required to match these characters.
- Otherwise, set `dp[i][j]` to the minimum of the following three options:
- `dp[i-1][j] + 1`: Delete the character at position `i` from `word1`.
- `dp[i][j-1] + 1`: Insert the character at position `j` from `word2` into `word1`.
- `dp[i-1][j-1] + 1`: Replace the character at position `i` in `word1` with the character at position `j` in `word2`.
6. Return `dp[m][n]`, which represents the minimum number of operations required to convert `word1` to `word2`.
Here's the implementation of the `minDistance` method in Java:
```java
class Solution {
public int minDistance(String word1, String word2) {
int m = word1.length();
int n = word2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 0; i <= m; i++) {
dp[i][0] = i;
}
for (int j = 0; j <= n; j++) {
dp[0][j] = j;
}
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = Math.min(dp[i - 1][j], Math.min(dp[i][j - 1], dp[i - 1][j - 1])) + 1;
}
}
}
return dp[m][n];
}
}
```
This implementation efficiently calculates the minimum edit distance between two strings using dynamic programming, with a time complexity of O(m * n) and a space complexity of O(m * n), where m is the length of `word1` and n is the length of `word2`.
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