GV

GeezerVison

The World Wide Web, restored to 800 pixels of glory.

Home|About|Gaming cabinet|Archives|How it works|Starter sites

On this site

  • Sign in
  • ActionsCI/CD and automation
  • CodespacesCloud development environments
  • All Features
  • Documentation
  • Training
  • Certifications
  • Build with Copilot
  • Customize Copilot
  • Integrations
  • Trending repos
  • GitHub Sponsors
  • Developer articles
  • Enterprises

This view

  • Original page
  • New address

GeezerVison is showing github.com — JavaScript/Dynamic-Programming/EditDistance.js at master · TheAlgorithms/JavaScript

Expand file tree

/

EditDistance.js

Copy path
More file actions
More file actions

Latest commit

 

History

History
History
54 lines (43 loc) · 1.31 KB
/

EditDistance.js

Copy path

File metadata and controls

54 lines (43 loc) · 1.31 KB
Raw
Copy raw file
Download raw file
Open symbols panel
Edit and raw actions
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
/*
Wikipedia -> https://en.wikipedia.org/wiki/Edit_distance
Q. -> Given two strings `word1` and `word2`. You can perform these operations on any of the string to make both strings similar.
- Insert
- Remove
- Replace
Find the minimum operation cost required to make both same. Each operation cost is 1.
Algorithm details ->
time complexity - O(n*m)
space complexity - O(n*m)
*/
const minimumEditDistance = (word1, word2) => {
const n = word1.length
const m = word2.length
const dp = new Array(m + 1).fill(0).map((item) => [])
/*
fill dp matrix with default values -
- first row is filled considering no elements in word2.
- first column filled considering no elements in word1.
*/
for (let i = 0; i < n + 1; i++) {
dp[0][i] = i
}
for (let i = 0; i < m + 1; i++) {
dp[i][0] = i
}
/*
indexing is 1 based for dp matrix as we defined some known values at first row and first column/
*/
for (let i = 1; i < m + 1; i++) {
for (let j = 1; j < n + 1; j++) {
const letter1 = word1[j - 1]
const letter2 = word2[i - 1]
if (letter1 === letter2) {
dp[i][j] = dp[i - 1][j - 1]
} else {
dp[i][j] = Math.min(dp[i - 1][j], dp[i - 1][j - 1], dp[i][j - 1]) + 1
}
}
}
return dp[m][n]
}
export { minimumEditDistance }
You are visitor 01402503 — GeezerVison does not store the pages you look up.
Best viewed with Netscape Navigator 4.7 or Internet Explorer 6 at 800×600. No cookies required.