# 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.

• 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];
}
}

• 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];
}
};

• class Solution:
def minDistance(self, word1: str, word2: str) -> int:
m, n = len(word1), len(word2)
dp = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(m + 1):
dp[i][0] = i
for j in range(n + 1):
dp[0][j] = j
for i in range(1, m + 1):
for j in range(1, n + 1):
if word1[i - 1] == word2[j - 1]:
dp[i][j] = dp[i - 1][j - 1]
else:
# dp[i - 1][j - 1]) meaning replace.
# e.g. "abc" and "bf", checking "ab" and "b" distance,
#   then replace either way for "c" or "f"
dp[i][j] = 1 + min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1])
return dp[-1][-1]

############

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 # merged for loops from above solution, but not good for readability
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]


• 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]
}

• 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];
}


• /**
* @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];
};