119. Edit Distance

119. Edit Distance

Tag:

Dynamic Problem

Main Idea:

DP-Match Problem.

1. State:
   dp[i][j] : the minimum number of steps of converting s1(0,i) to s2(0, j).
2. Update Function:
   IF s1[i] == s2[j], THEN dp[i][j] = min(dp[i-1][j-1], 1+min(dp[i-1][j], dp[i][j-1]));
   ELSE, dp[i][j] = 1 + min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]).
3. Initialization:
   1. dp[i][0] =i
   2. dp[0][j] =j
4. Answer:
   dp[n][m]

Tips/Notes:

Time/Space Cost:

Time Cost: $\ O(nm)$
Space Cost: $\ O(nm)$

Code:

class Solution {
public:
    /**
     * @param word1: A string
     * @param word2: A string
     * @return: The minimum number of steps.
     */
    int minDistance(string &word1, string &word2) {
        // write your code here
        int n = word1.length();
        int m = word2.length();
        
        vector<vector<int>> dp(n+1, vector<int>(m+1, 0));
        
        for(int i = 0; i <= n; i++){
            dp[i][0] = i;
        }
        
        for(int j = 0; j <= m; j++){
            dp[0][j] = j;
        }
        
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                if(word1[i-1] == word2[j-1])
                    dp[i][j] = min(dp[i-1][j-1], 1 + min(dp[i-1][j], dp[i][j-1]));
                    
                else
                    dp[i][j] = 1 + min(dp[i-1][j-1], min(dp[i-1][j], dp[i][j-1]));
            }
        }
        
        return dp[n][m];
        
    }
};