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

Analysis:b) Delete a character

c) Replace a character

At the first glance, we might think this is a DFS problem, but if we see it is hard to find a quick DFS thought and the problem requires some optimal result (here is the minimum), DP is a good direction to consider.

Actually this is a classic DP problem:

The state is: table[i][j]=minimum number steps convert word1[1:i] to word2[1:j] (here assume string starts from 1).

The optimal function is: table[i+1][j+1] = min [table[i][j]+1 or 0 (+0 if word1[i+1]==word2[j+1], else +1), table[i][j+1]+1, table[i+1][j]+1 ].

Initialization:

table[0][i] = i i=1:|word1| here 0 means "", any string convert to "" needs the length of string

table[j][0] = j i=1:|word2|

table[0][0]=0 "" convert to "" need 0 steps.

Code(updated 201401):

class Solution { public: int minDistance(string word1, string word2) { int s1 = word1.size(); int s2 = word2.size(); if (s1==0){return s2;} if (s2==0){return s1;} vector<vector<int> > w(s1+1,vector<int>(s2+1,0)); for (int i=0;i<=s1;i++){w[i][0]=i;} for (int i=0;i<=s2;i++){w[0][i]=i;} for (int i=1;i<=s1;i++){ for (int j=1;j<=s2;j++){ w[i][j]=min(w[i-1][j]+1,w[i][j-1]+1); if (word1[i-1]==word2[j-1]){ w[i][j]=min(w[i-1][j-1],w[i][j]); }else{ w[i][j]=min(w[i-1][j-1]+1,w[i][j]); } } } return w[s1][s2]; } };

Code:

class Solution { public: int minDistance(string word1, string word2) { // Start typing your C/C++ solution below // DO NOT write int main() function int len1 = word1.size(); int len2 = word2.size(); vector<vector<int> > table(len1+1,vector<int>(len2+1,0)); for (int i=1;i<=len1;i++){ table[i][0]=i; } for (int i=1;i<=len2;i++){ table[0][i]=i; } for(int i=0;i<len1;i++){ for (int j=0;j<len2;j++){ table[i+1][j+1]=min(table[i+1][j],table[i][j+1])+1; if (word1[i]==word2[j]){ table[i+1][j+1]=min(table[i+1][j+1],table[i][j]); }else{ table[i+1][j+1]=min(table[i+1][j+1],table[i][j]+1); } } } return table[len1][len2]; } };

A nice solution!

ReplyDeleteI think there is a typo: It should be

The optimal function is: table[i+1][j+1] = min [table[i][j]+1 or 0

thanks for your reply ! I have corrected the typo in the post.

DeleteNice one. But I may find a typo for you as well.

ReplyDeleteFor this one: table[i][j]+1 or 0 (+1 if word1[i+1]==word2[j+1], else +0). I think it should be if word1[i+1] == word2[j+1], we should do +0, else we will do +1. This should make your explanation correspond to the codes

table[i+1][j+1] = min [table[i][j]+1 or 0 (+1 if word1[i+1]==word2[j+1], else +0),

ReplyDeleteI believe you should switch +0 and +1....

Excellent dp solution..very helpful..how does it strikes this algo can strike in one's mind?

ReplyDelete