Baozi Leetcode solution 72. Edit Distance

Problem Statement

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

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 link

Video Tutorial

You can find the detailed video tutorial here

• Youtube
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Thought Process

Brute force sounds like really overkill because we have to list all the possible ways and it is exponential time complexity. On a high level, the recursion looks like

• if cannot convert A to B, return -1 (the recursion termination function); if A == B, then return 0
• try 3 different ways, insert, remove and replace, cut that character and continue the recursion. Compare the minimum of that 3 methods (exclude -1)

Edit distance is a classic Dynamic Programming (DP) problem, just like Coin Change Problem. It follows the DP template perfectly (asking for extreme values)

The mathematical induction function is

DP[i][j] is the minimum operations needed to convert String[0-i] to String[0-j]

Initial values: dp[0][j] = j and dp[i][0] = i

If (A[i] == B[j]) dp[i][j] =  dp[i-1][j-1]

Else min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) +1;

Solutions

DP

 1 public int minDistance(String word1, String word2) {
 2     if (word1 == null || word2 == null) return -1;
 3
 4     int l1 = word1.length();
 5     int l2 = word2.length();
 6
 7     if (l1 == 0 || l2 == 0) {
 8         return l1 == 0 ? l2 : l1;
 9     }
10
11     // A pattern for coding up DP, use extra array
12     // This could be further optimized into 2 arrays
13     int[][] lookup = new int[l1 + 1][l2 + 1];
14
15     // Note the initialization case here, for an empty string, it will take i to change with the current one
16     // initialize
17     for (int i = 0; i <= l2; i++) {
18         lookup[0][i] = i;
19     }
20
21     for (int i = 0; i <= l1;i++) {
22         lookup[i][0] = i;
23     }
24
25     /* mathematical induction function:
26      * dp[i][j] =  dp[i-1][j-1]   if (A[i] == B[j])
27        or = min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) +1;
28         dp[0][j] = j and dp[i][0] = i   */
29     // this is O(M*N) time and M*N space, space could be saved using 2 arrays
30     for (int i = 1; i <= l1; i++) {
31         for (int j = 1; j <= l2; j++) {
32             if (word1.charAt(i-1) == word2.charAt(j-1)) {
33                 lookup[i][j] = lookup[i-1][j-1];
34             } else {
35                 lookup[i][j] = Math.min(lookup[i-1][j], Math.min(lookup[i][j-1], lookup[i-1][j-1])) + 1;
36             }
37         }
38     }
39
40     return lookup[l1][l2];
41 }

Time Complexity: O(M*N) where M is word1 length and N is word2 length

Space Complexity: O(M*N) since we need an extra 2D array

References

• Leetcode official solution

原文地址：https://www.cnblogs.com/baozitraining/p/12114034.html