72. Edit Distance

Description

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:

1. Insert a character

2. Delete a character

3. Replace a character

Constraints

Approach

Levenshtein distance

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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')

Solutions

Solution 1

/**
 * Time complexity : O(mn) as it follows quite straightforward for the inserted loops.
 * Space complexity : O(mn) since at each step we keep the results of all 
 *    previous computations.
 */

class Solution {
    public int minDistance(String word1, String word2) {
        int w1Len = word1.length();
        int w2Len = word2.length();
        // if one of the strings is empty
        if(w1Len * w2Len == 0) return w1Len + w2Len;
        
        int[][] dp = new int[w2Len+1][w1Len+1];
        // init boundaries
        for(int i = 0; i <= w1Len; i++) {
            dp[0][i] = i;
        }
        for(int i = 0; i <= w2Len; i++) {
            dp[i][0] = i;
        }
        // DP compute
        for(int i = 1; i <= w2Len; i++) {
            for(int j = 1; j <= w1Len; j++) {
                if(word2.charAt(i-1) == word1.charAt(j-1)) {
                    dp[i][j] = dp[i-1][j-1];
                } else {
                    dp[i][j] = Math.min(dp[i-1][j-1], 
                                    Math.min(dp[i-1][j], dp[i][j-1])) + 1;
                }
            }
        }
        return dp[w2Len][w1Len];
    }
}

Solution 2

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    int[][] dp;
    public int minDistance(String a, String b) {
        int n = a.length(), m = b.length();
        dp = new int[a.length()][b.length()];
        return f(n-1, m-1, a, b);
    }
    public int f(int i, int j, String a, String b) {
        if (i < 0) return j+1;
        if (j < 0) return i+1;
        if (dp[i][j] != 0) return dp[i][j];
        if (a.charAt(i) == b.charAt(j)) {
            dp[i][j] = f(i-1, j-1, a, b);
        } else {
            dp[i][j] = 1 + Math.min(f(i-1, j-1, a, b) , 
                            Math.min(f(i-1, j, a, b), f(i, j-1, a, b)));
        }
        return dp[i][j];
    }
}

Follow up

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