Edit Distance — DSA-30

Edit Distance hard

Description

Given two strings word1 and word2, return the minimum number of operations required to convert word1 into word2.

You have three operations:

Insert a character
Delete a character
Replace a character

Also known as Levenshtein distance. The bedrock of fuzzy matching, spell check, DNA alignment, and git diff.

Examples

> Case 1:
    word1 = "horse",  word2 = "ros"
    Output: 3
    Explanation:
        horse -> rorse  (replace 'h' with 'r')
        rorse -> rose   (delete 'r')
        rose  -> ros    (delete 'e')
 
> Case 2:
    word1 = "intention",  word2 = "execution"
    Output: 5

Constraints

0 <= word1.length, word2.length <= 500

State design

What am I computing? dp[i][j] = min edits to transform word1[0..i) into word2[0..j).
Dimensions? Two.
Base cases? dp[0][j] = j (turn empty into word2[0..j) by inserting j chars). dp[i][0] = i (turn word1[0..i) into empty by deleting i chars).
Transition?
- Match (word1[i - 1] == word2[j - 1]): no edit needed. dp[i][j] = dp[i - 1][j - 1].
- Mismatch: try all three operations, take the min, add 1:
  - Replace the last char: dp[i - 1][j - 1] + 1.
  - Delete the last char of word1: dp[i - 1][j] + 1.
  - Insert a char to match word2[j - 1]: dp[i][j - 1] + 1.
Order? Increasing i, increasing j.
Answer? dp[n][m].

The three transitions correspond to the three legal moves on the “alignment grid”:

Diagonal = consume one char from each side (match → free; mismatch → +1 replace).
Down = consume one char from word1 only (= delete it).
Right = consume one char from word2 only (= insert it).

Edit distance is the shortest path through this grid where every step costs at most 1.

Try it yourself

⌨ Edit Distance — return the minimum number of edits

def min_distance(word1, word2):
  # dp[i][j] = edits to turn word1[:i] into word2[:j]; match is free, else 1 + min of three moves
  pass

Hint

Base cases: dp[i][0] = i, dp[0][j] = j. If the last chars match, dp[i-1][j-1]; else 1 + min(replace=dp[i-1][j-1], delete=dp[i-1][j], insert=dp[i][j-1]).

Reveal solution

Code (2D table)

Space-optimized to O(m)

Analysis

Time: O(n × m).
Space: O(n × m) for the 2D table, O(m) rolled.

Same skin

Delete Operation for Two Strings — only delete allowed; equivalently n + m − 2 × LCS.
Minimum ASCII Delete Sum for Two Strings — same recurrence, edit cost = ASCII value of deleted char instead of 1.
One Edit Distance — early exit when count exceeds 1.
Wildcard / Regex Matching — extend the transition with * and ? (or * and .).

The Levenshtein recurrence is the spine of most string-DP problems — once it’s in your fingers, an entire family of problems takes the same shape.

Maximum Product Subarray Overview

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