LeetCode-72: Edit Distance

topic:

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

Insert a character
Delete a character
Replace a character

example

Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation: 
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example 2:

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation: 
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

Problem analysis:

Find the minimum edit distance of two strings.

Link:

• https://leetcode.com/problems/edit-distance/description/

Thought tags:

Algorithm: Dynamic Programming

answer:

• Use dp[i][j] to represent the minimum edit distance of 0~i-1 of string 1 and 0~j-1 of string 2;
• We can know the edge cases: dp[i][0] = i, dp[0][j]=j;
• At the same time, for the substrings of two strings, it can be divided into the cases where the last character is equal or unequal:
• If words[i-1] == words[j-1]: dp[i][j] = dp[i-1][j-1]; that is to say, the current edit distance has nothing to do with the characters at positions i and j;
• If words[i-1] != words[j-1]: then there are three possible actions:
  
  • Insert into word1: dp[i][j] = dp[i][j-1] + 1;
  • Delete from word1: dp[i][j] = dp[i-1][j] + 1;
  • Replace word1 element: dp[i][j] = dp[i-1][j-1] + 1;

class Solution {
public:
    int minDistance(string word1, string word2) {
        int rows = word1.length();
        int cols = word2.length();
        vector<vector<int> > dp(rows+1, vector<int>(cols+1, 0));

        for(int i=1; i<=rows; ++i)
            dp[i][0] = i;
        for(int j=1; j<=cols; ++j)
            dp[0][j] = j;

        for(int i=1; i<=rows; ++i){
            for(int j=1; j<=cols; ++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], min(dp[i-1][j], dp[i][j-1])) + 1;
            }
        }

        return dp[rows][cols];
    }
};