LeetCode: Edit Distance

Edit Distance



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

Using DP, time requirement is O(mn), storage can be O(mn) or O(min(m,n)). See two solutions below. 

public class Solution {
    public int minDistance(String word1, String word2) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int m = word1.length(), n = word2.length();
        int[][] distances = new int[m+1][n+1];
        distances[0][0] = 0;
        for(int i=1;i<=m;i++)
            distances[i][0]=i;
        for(int i=1;i<=n;i++)
            distances[0][i]=i;
        for(int i=1;i<=m;i++){
            for(int j=1;j<=n;j++){
                if(word1.charAt(i-1)==word2.charAt(j-1))
                    distances[i][j] = distances[i-1][j-1];
                else
                    distances[i][j] = 1+Math.min(distances[i-1][j-1],Math.min(distances[i-1][j],distances[i][j-1]));
            }
        }
        return distances[m][n];
    }
}

public class Solution {
    public int minDistance(String word1, String word2) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int m = word1.length(), n = word2.length();
        if(m==0) return n;
        if(n==0) return m;
        
        int[] distances = new int[n+1],old = new int[n+1]; 
        for(int i=0;i<=n;i++)
            old[i]=i;

        
        for(int i=1;i<=m;i++){
            for(int j=1;j<=n;j++){
                if(word1.charAt(i-1)==word2.charAt(j-1))
                    distances[j] = j>1?old[j-1]:(i-1);
                else
                    distances[j] = 1+Math.min(j>1?old[j-1]:(i-1),Math.min(old[j],j>1?distances[j-1]:i));
            }
            for(int j=1;j<=n;j++)
                old[j]=distances[j];
        }
        return old[n];
    }
}