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.

Solution

function minDistance(word1: string, word2: string): number {
    const memo: number[][] = new Array(word1.length + 1).fill(0).map((_) => [])
    const l1 = word1.length
    const l2 = word2.length
    const dfs = (w1: number, w2: number): number => {
        if (memo[w1][w2] != undefined) return memo[w1][w2]
        if (w1 == l1 && w2 == l2) {
            memo[w1][w2] = 0
            return 0
        }
        if (w1 == l1 || w2 == l2) {
            memo[w1][w2] = Math.max(l1 - w1, l2 - w2)
            return memo[w1][w2]
        }
        let result = 0
        if (word1[w1] == word2[w2]) {
            result = dfs(w1 + 1, w2 + 1)
        } else {
            result = 1 + Math.min(dfs(w1 + 1, w2), dfs(w1 + 1, w2 + 1), dfs(w1, w2 + 1))
        }
        memo[w1][w2] = result
        return result
    }
    return dfs(0, 0)
}

export { minDistance }

This site is open source. Improve this page.