## Edit Distance

• Time:O(mn)
• Space:O(mn) \to O(n)

## C++

class Solution {
public:
int minDistance(string word1, string word2) {
const int m = word1.length();
const int n = word2.length();
// dp[i][j] := min # of operations to convert word1[0..i) to word2[0..j)
vector<vector<int>> dp(m + 1, vector<int>(n + 1));

for (int i = 1; i <= m; ++i)
dp[i][0] = i;

for (int j = 1; j <= n; ++j)
dp[0][j] = j;

for (int i = 1; i <= m; ++i)
for (int j = 1; j <= n; ++j)
if (word1[i - 1] == word2[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[m][n];
}
};


## JAVA

class Solution {
public int minDistance(String word1, String word2) {
final int m = word1.length();
final int n = word2.length();
// dp[i][j] := min # of operations to convert word1[0..i) to word2[0..j)
int[][] dp = new int[m + 1][n + 1];

for (int i = 1; i <= m; ++i)
dp[i][0] = i;

for (int j = 1; j <= n; ++j)
dp[0][j] = j;

for (int i = 1; i <= m; ++i)
for (int j = 1; j <= n; ++j)
if (word1.charAt(i - 1) == word2.charAt(j - 1))
dp[i][j] = dp[i - 1][j - 1];
else
dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;

return dp[m][n];
}
}


## Python

class Solution:
def minDistance(self, word1: str, word2: str) -> int:
m = len(word1)
n = len(word2)
# dp[i][j] := min # of operations to convert word1[0..i) to word2[0..j)
dp = [[0] * (n + 1) for _ in range(m + 1)]

for i in range(1, m + 1):
dp[i][0] = i

for j in range(1, 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][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1

return dp[m][n]