# 72. Edit Distance

### Description

Given two words *word1* and *word2*, find the minimum number of operations required to convert *word1* to *word2*.

You have the following 3 operations permitted on a word:

1. Insert a character
2. Delete a character
3. Replace a character

### Constraints

### Approach

[Levenshtein distance](https://en.wikipedia.org/wiki/Levenshtein_distance)

### Links

* [GeeksforGeeks](https://www.geeksforgeeks.org/edit-distance-dp-5)
* [Leetcode](https://leetcode.com/problems/edit-distance)
* ProgramCreek
* [YouTube](https://youtu.be/We3YDTzNXEk)

### Examples

{% tabs %}
{% tab title="Example 1" %}
**Input:** word1 = "horse", word2 = "ros"

**Output:** 3

**Explanation:**

horse -> rorse (replace 'h' with 'r')

rorse -> rose (remove 'r')

rose -> ros (remove 'e')
{% endtab %}

{% tab title="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')
{% endtab %}
{% endtabs %}

### Solutions

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : O(mn) as it follows quite straightforward for the inserted loops.
 * Space complexity : O(mn) since at each step we keep the results of all 
 *    previous computations.
 */

class Solution {
    public int minDistance(String word1, String word2) {
        int w1Len = word1.length();
        int w2Len = word2.length();
        // if one of the strings is empty
        if(w1Len * w2Len == 0) return w1Len + w2Len;
        
        int[][] dp = new int[w2Len+1][w1Len+1];
        // init boundaries
        for(int i = 0; i <= w1Len; i++) {
            dp[0][i] = i;
        }
        for(int i = 0; i <= w2Len; i++) {
            dp[i][0] = i;
        }
        // DP compute
        for(int i = 1; i <= w2Len; i++) {
            for(int j = 1; j <= w1Len; j++) {
                if(word2.charAt(i-1) == word1.charAt(j-1)) {
                    dp[i][j] = dp[i-1][j-1];
                } else {
                    dp[i][j] = Math.min(dp[i-1][j-1], 
                                    Math.min(dp[i-1][j], dp[i][j-1])) + 1;
                }
            }
        }
        return dp[w2Len][w1Len];
    }
}
```

{% endtab %}

{% tab title="Solution 2" %}

```java
/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    int[][] dp;
    public int minDistance(String a, String b) {
        int n = a.length(), m = b.length();
        dp = new int[a.length()][b.length()];
        return f(n-1, m-1, a, b);
    }
    public int f(int i, int j, String a, String b) {
        if (i < 0) return j+1;
        if (j < 0) return i+1;
        if (dp[i][j] != 0) return dp[i][j];
        if (a.charAt(i) == b.charAt(j)) {
            dp[i][j] = f(i-1, j-1, a, b);
        } else {
            dp[i][j] = 1 + Math.min(f(i-1, j-1, a, b) , 
                            Math.min(f(i-1, j, a, b), f(i, j-1, a, b)));
        }
        return dp[i][j];
    }
}
```

{% endtab %}
{% endtabs %}

### Follow up

* Given two words *word1* and *word2*, **find** and **print** the minimum number of operations required to convert *word1* to *word2*.


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://code-snippets.hbamithkumara.com/leetcode/problems/1-100/edit-distance.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.