[LeetCode] Edit Distance

This is a classic problem of Dynamic Programming. We define the state dp[i][j] to be the minimum number of operations to convert word1[0..i - 1] to word2[0..j - 1]. The state equations have two cases: the boundary case and the general case. Note that in the above notations, both i and j take values starting from 1.

For the boundary case, that is, to convert a string to an empty string, it is easy to see that the mininum number of operations to convert word1[0..i - 1] to "" requires at least i operations (deletions). In fact, the boundary case is simply:

1. dp[i][0] = i;
2. dp[0][j] = j.

Now let's move on to the general case, that is, convert a non-empty word1[0..i - 1] to another non-empty word2[0..j - 1]. Well, let's try to break this problem down into smaller problems (sub-problems). Suppose we have already known how to convert word1[0..i - 2] to word2[0..j - 2], which is dp[i - 1][j - 1]. Now let's consider word[i - 1] and word2[j - 1]. If they are euqal, then no more operation is needed and dp[i][j] = dp[i - 1][j - 1]. Well, what if they are not equal?

If they are not equal, we need to consider three cases:

1. Replace word1[i - 1] by word2[j - 1] (dp[i][j] = dp[i - 1][j - 1] + 1 (for replacement));
2. Delete word1[i - 1] and word1[0..i - 2] = word2[0..j - 1] (dp[i][j] = dp[i - 1][j] + 1 (for deletion));
3. Insert word2[j - 1] to word1[0..i - 1] and word1[0..i - 1] + word2[j - 1] = word2[0..j - 1] (dp[i][j] = dp[i][j - 1] + 1 (for insertion)).

Make sure you understand the subtle differences between the equations for deletion and insertion. For deletion, we are actually converting word1[0..i - 2] to word2[0..j - 1], which costs dp[i - 1][j], and then deleting the word1[i - 1], which costs 1. The case is similar for insertion.

Putting these together, we now have:

1. dp[i][0] = i;
2. dp[0][j] = j;
3. dp[i][j] = dp[i - 1][j - 1], if word1[i - 1] = word2[j - 1];
4. dp[i][j] = min(dp[i - 1][j - 1] + 1, dp[i - 1][j] + 1, dp[i][j - 1] + 1), otherwise.

The above state equations can be turned into the following code directly.

class Solution { 
public:
    int minDistance(string word1, string word2) { 
        int m = word1.length(), n = word2.length();
        vector<vector<int> > dp(m + 1, vector<int> (n + 1, 0));
        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] + 1, min(dp[i][j - 1] + 1, dp[i - 1][j] + 1));
            }
        }
        return dp[m][n];
    }
};

Well, you may have noticed that each time when we update dp[i][j], we only need dp[i - 1][j - 1], dp[i][j - 1], dp[i - 1][j]. In fact, we need not maintain the full m*n matrix. Instead, maintaing one column is enough. The code can be optimized to O(m) or O(n) space, depending on whether you maintain a row or a column of the original matrix.

The optimized code is as follows.

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

Well, if you find the above code hard to understand, you may first try to write a two-column version that explicitly maintains two columns (the previous column and the current column) and then simplify the two-column version into the one-column version like the above code :-)