Leetcode-Edit Distance(dp)

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

DISCUSS:

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 considerword[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.

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