[Leetcode] Edit Distance

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

For the boundary case, that's, to convert a string to an empty string, it's easy-to-see that the mininum number of opera tions 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's, convert a non-empty to word1[0..i - 1] another non-empty word2[0..j - 1] . Well, let's try-to-break this problem-to-smaller problems (sub-problems). Suppose we have already known how to convert word1[0..i - 2] word2[0..j - 2] to, which is dp[i - 1][j - 1] . Now let ' s consider word[i - 1] and word2[j - 1] . If They is euqal, then no more operation is needed and dp[i][j] = dp[i - 1][j - 1] . Well, what if they is not equal?

If They is 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 to and word2[j - 1] word1[0..i - 1] word1[0..i - 1] + word2[j - 1] = word2[0..j - 1] ( dp[i][j] = dp[i][j - 1] + 1 (for insertion) ).

Make sure your understand the subtle differences between the equations for deletion and insertion. For deletion, we is 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 c15/>. The case was 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 is turned into the following code directly.

1 classSolution {2  Public:3     intMindistance (stringWord1,stringWord2) { 4         intm = Word1.length (), n =word2.length ();5vector<vector<int> > dp (M +1, vector<int> (n +1,0));6          for(inti =1; I <= m; i++)7dp[i][0] =i;8          for(intj =1; J <= N; J + +)9dp[0][J] =J; Ten          for(inti =1; I <= m; i++) { One              for(intj =1; J <= N; J + +) { A                 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][j-1] +1, Dp[i-1][J] +1)); the             } -         } -         returnDp[m][n]; -     } +};

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

The optimized code is as follows.

1 classSolution {2  Public:3     intMindistance (stringWord1,stringWord2) {4         intm = Word1.length (), n =word2.length ();5vector<int> cur (m +1,0);6          for(inti =1; I <= m; i++)7Cur[i] =i;8          for(intj =1; J <= N; J + +) {9             intPre = cur[0];Tencur[0] =J; One              for(inti =1; I <= m; i++) { A                 inttemp =Cur[i]; -                 if(Word1[i-1] = = Word2[j-1]) -Cur[i] =Pre; the                 ElseCur[i] = min (pre +1, Min (Cur[i] +1, Cur[i-1] +1)); -Pre =temp; -             } -         } +         returnCur[m]; -     } +};

Well, if you find the above code hard-understand, you could first try to write a two-column version of that explicitly maint Ains columns (the previous column and the current column) and then simplify the Two-column version into the One-column Version like the above code:-)

[Leetcode] Edit Distance