for "sea" and "eat"

current id i and j

if word1[i] == word[j], then dp[i][j] = dp[i - 1][j - 1];

else

delete word1[i],then dp = dp[i - 1][j] + 1;

delete word2[j],then dp = dp[i][j - 1] + 1;

so dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1;

'''

class Solution {

public:

int minDistance(string word1, string word2) {

int len1 = word1.size();

int len2 = word2.size();

vector<vector<int>> dp(len1 + 1, vector<int>(len2 + 1, 0));

```
for (int i = 1; i <= len1; i++)
dp[i][0] = i;
for (int j = 1; j <= len2; j++)
dp[0][j] = j;
for (int i = 1; i <= len1; i++)
{
for (int j = 1; j <= len2; 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], dp[i][j - 1]) + 1;
}
}
return dp[len1][len2];
}
```

};

'''