LeetCode-in-All

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 Swift 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 Swift:

class Solution {
    func minDistance(_ word1: String, _ word2: String) -> Int {
        var n = word1.count
        var m = word2.count
        if word1 == word2 { return 0}
        if m == 0 && n == 0 { return 0 }
        if m == 1 || n == 1 { return 1}
        var word1 = Array(word1)
        var word2 = Array(word2)
        var dp: [[Int]] = Array(repeating:Array(repeating: 0, count:m+1), count:n+1)
        for i in 0...n {
            dp[i][0] = i
        }
        for j in 0...m {
            dp[0][j] = j
        }
        for i in 1...n {
            for j in 1...m {
                if word1[i-1] == word2[j-1] {
                     dp[i][j] = dp[i-1][j-1]
                } else {
                     dp[i][j] = min(min(dp[i-1][j-1], dp[i-1][j]), dp[i][j-1]) + 1
                }
            }
        }
        return dp[n][m]
    }
}

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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