LeetCode - #72 Edit Distance (Top 100)

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Difficulty Level: Difficult

1. Description

Given two words word1and , please return the minimum number of operands used word2to word1convert to .word2

You can perform the following three operations on a word:

• insert a character
• delete a character
• replace a character

2. Example

Example 1

输入：word1 = "horse", word2 = "ros"
输出：3
解释：
horse -> rorse (将 'h' 替换为 'r')
rorse -> rose (删除 'r')
rose -> ros (删除 'e')
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Example 2

输入：word1 = "intention", word2 = "execution"
输出：5
解释：
intention -> inention (删除 't')
inention -> enention (将 'i' 替换为 'e')
enention -> exention (将 'n' 替换为 'x')
exention -> exection (将 'n' 替换为 'c')
exection -> execution (插入 'u')
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Restrictions:

• 0 <= word1.length, word2.length <= 500
• word1and word2consists of lowercase English letters

3. Answers

class EditDistance {
    func minDistance(word1: String, _ word2: String) -> Int {
        let aChars = [Character](word1.characters)
        let bChars = [Character](word2.characters)
        let aLen = aChars.count
        let bLen = bChars.count
        
        var dp = Array(count: aLen + 1, repeatedValue:(Array(count: bLen + 1, repeatedValue: 0)))
        
        for i in 0...aLen {
            for j in 0...bLen {
                if i == 0 {
                    dp[i][j] = j
                } else if j == 0 {
                    dp[i][j] = i
                } else if aChars[i - 1] == bChars[j - 1] {
                    dp[i][j] = dp[i - 1][j - 1]
                } else {
                    dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1
                }
            }
        }
        
        return dp[aLen][bLen]
    }
}
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• Main idea: 2D dynamic programming, find the smallest steps from inserting, deleting or replacing a character.
• Time Complexity: O(mn)
• Space Complexity: O(mn)

Repository for the algorithm solution: LeetCode-Swift

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