13.9 Edit Distance

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Link: https://oj.leetcode.com/problems/edit-distance/

Below is my 1st attempt. But has errors.

public class Solution {    public int minDistance(String word1, String word2) {        if(word1 == null && word2 == null) return 0;        int m = word1.length();        int n = word2.length();        int[][] f = new int[m][n];        for(int i = 0; i < m; i++){            for(int j = 0; j < n; j++){                int score = word1.charAt(i) == word2.charAt(j) ? 1 : 0;                f[i][j] = Math.min(Math.min(f[i-1][j], f[i][j-1])+1, f[i-1][j-1] + score);            }        }        return f[m-1][n-1];    }}

Approach I: correct answer

Time: O(mn), Space: O(mn)

public class Solution {    public int minDistance(String word1, String word2) {        if(word1 == null && word2 == null) return 0;        int m = word1.length();        int n = word2.length();        int[][] f = new int[m+1][n+1];        //init the first row        for(int j = 0; j <=n; j++){            f[0][j] = j;        }        //init the first col        for(int i = 0; i <=m; i++){            f[i][0] = i;        }        //fill in the middle        for(int i = 1; i <=m; i++){            for(int j = 1; j <=n; j++){                if(word1.charAt(i-1) == word2.charAt(j-1)){                    f[i][j] = f[i-1][j-1];//why don't need to compare with Math.min(f[i-1][j], f[i][j-1])+1 here?                }                 else{                    f[i][j] = Math.min(Math.min(f[i-1][j], f[i][j-1]), f[i-1][j-1])+1;                }            }        }        return f[m][n];    }}

Approach II: DP with two arrays

Since f[i][j] come from f[i-1][j-1], f[i][j-1] and f[i-1][j], we cannot use one array to store it. So we create two arrays. One to store the current row, the other to store the previous row.

Time: O(mn), Space: O(min(m, n))

public class Solution {    public int minDistance(String word1, String word2) {        if(word1 == null && word2 == null) return 0;        int m = word1.length();        int n = word2.length();        int[] f = new int[n+1];        //init the first row        for(int j = 0; j <=n; j++){            f[j] = j;        }        //fill in the middle        for(int i = 1; i <=m; i++){            int[] newF = new int[n+1];            newF[0] = i;            for(int j = 1; j <=n; j++){                if(word1.charAt(i-1) == word2.charAt(j-1)){                    newF[j] = f[j-1];                }                 else{                    newF[j] = Math.min(Math.min(f[j], newF[j-1]), f[j-1])+1;                }            }            f = newF;        }        return f[n];    }}

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