72. Edit Distance

Description

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.

Solutions

Solution 1: Dynamic Programming

We define $f[i][j]$ as the minimum number of operations to convert $word1$ of length $i$ to $word2$ of length $j$ . $f[i][0] = i$ , $f[0][j] = j$ , $i \in [1, m], j \in [0, n]$ .

We consider $f[i][j]$ :

If $word1[i - 1] = word2[j - 1]$ , then we only need to consider the minimum number of operations to convert $word1$ of length $i - 1$ to $word2$ of length $j - 1$ , so $f[i][j] = f[i - 1][j - 1]$ ;
Otherwise, we can consider insert, delete, and replace operations, then $f[i][j] = \min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1$ .

Finally, we can get the state transition equation:

$f[i][j] = \begin{cases} i, & \text{if } j = 0 \ j, & \text{if } i = 0 \ f[i - 1][j - 1], & \text{if } word1[i - 1] = word2[j - 1] \ \min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1, & \text{otherwise} \end{cases}$

Finally, we return $f[m][n]$ .

The time complexity is $O(m \times n)$ , and the space complexity is $O(m \times n)$ . $m$ and $n$ are the lengths of $word1$ and $word2$ respectively.

Python Code

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class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)
        f = [[0] * (n + 1) for _ in range(m + 1)]
        for j in range(1, n + 1):
            f[0][j] = j
        for i, a in enumerate(word1, 1):
            f[i][0] = i
            for j, b in enumerate(word2, 1):
                if a == b:
                    f[i][j] = f[i - 1][j - 1]
                else:
                    f[i][j] = min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1
        return f[m][n]

Java Code

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class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length(), n = word2.length();
        int[][] f = new int[m + 1][n + 1];
        for (int j = 1; j <= n; ++j) {
            f[0][j] = j;
        }
        for (int i = 1; i <= m; ++i) {
            f[i][0] = i;
            for (int j = 1; j <= n; ++j) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    f[i][j] = f[i - 1][j - 1];
                } else {
                    f[i][j] = Math.min(f[i - 1][j], Math.min(f[i][j - 1], f[i - 1][j - 1])) + 1;
                }
            }
        }
        return f[m][n];
    }
}

C++ Code

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class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.size(), n = word2.size();
        int f[m + 1][n + 1];
        for (int j = 0; j <= n; ++j) {
            f[0][j] = j;
        }
        for (int i = 1; i <= m; ++i) {
            f[i][0] = i;
            for (int j = 1; j <= n; ++j) {
                if (word1[i - 1] == word2[j - 1]) {
                    f[i][j] = f[i - 1][j - 1];
                } else {
                    f[i][j] = min({f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]}) + 1;
                }
            }
        }
        return f[m][n];
    }
};

Go Code

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func minDistance(word1 string, word2 string) int {
	m, n := len(word1), len(word2)
	f := make([][]int, m+1)
	for i := range f {
		f[i] = make([]int, n+1)
	}
	for j := 1; j <= n; j++ {
		f[0][j] = j
	}
	for i := 1; i <= m; i++ {
		f[i][0] = i
		for j := 1; j <= n; j++ {
			if word1[i-1] == word2[j-1] {
				f[i][j] = f[i-1][j-1]
			} else {
				f[i][j] = min(f[i-1][j], min(f[i][j-1], f[i-1][j-1])) + 1
			}
		}
	}
	return f[m][n]
}

TypeScript Code

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function minDistance(word1: string, word2: string): number {
    const m = word1.length;
    const n = word2.length;
    const f: number[][] = Array(m + 1)
        .fill(0)
        .map(() => Array(n + 1).fill(0));
    for (let j = 1; j <= n; ++j) {
        f[0][j] = j;
    }
    for (let i = 1; i <= m; ++i) {
        f[i][0] = i;
        for (let j = 1; j <= n; ++j) {
            if (word1[i - 1] === word2[j - 1]) {
                f[i][j] = f[i - 1][j - 1];
            } else {
                f[i][j] = Math.min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1;
            }
        }
    }
    return f[m][n];
}

JavaScript Code

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/**
 * @param {string} word1
 * @param {string} word2
 * @return {number}
 */
var minDistance = function (word1, word2) {
    const m = word1.length;
    const n = word2.length;
    const f = Array(m + 1)
        .fill(0)
        .map(() => Array(n + 1).fill(0));
    for (let j = 1; j <= n; ++j) {
        f[0][j] = j;
    }
    for (let i = 1; i <= m; ++i) {
        f[i][0] = i;
        for (let j = 1; j <= n; ++j) {
            if (word1[i - 1] === word2[j - 1]) {
                f[i][j] = f[i - 1][j - 1];
            } else {
                f[i][j] = Math.min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1;
            }
        }
    }
    return f[m][n];
};

72. Edit Distance#

Description#

Solutions#

Solution 1: Dynamic Programming#

72. Edit Distance

Description

Solutions

Solution 1: Dynamic Programming