LeetCode in Racket

72. Edit Distance

Hard

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

You have the following three operations permitted on a word:

• Insert a character
• Delete a character
• 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’)

Constraints:

• 0 <= word1.length, word2.length <= 500
• word1 and word2 consist of lowercase English letters.

Solution

(define/contract (min-distance word1 word2)
  (-> string? string? exact-integer?)
  (let* ((n1 (string-length word1))
         (n2 (string-length word2)))
    (if (> n2 n1)
        (min-distance word2 word1)
        (let ((dp (make-vector (+ n2 1) 0)))
          (for ([j (in-range (+ n2 1))])
            (vector-set! dp j j))
          (for ([i (in-range 1 (+ n1 1))])
            (let ((pre (vector-ref dp 0)))
              (vector-set! dp 0 i)
              (for ([j (in-range 1 (+ n2 1))])
                (let* ((tmp (vector-ref dp j))
                       (cost (if (char=? (string-ref word1 (- i 1)) (string-ref word2 (- j 1)))
                                 pre
                                 (+ 1 (min pre (vector-ref dp j) (vector-ref dp (- j 1)))))))
                  (vector-set! dp j cost)
                  (set! pre tmp)))))
          (vector-ref dp n2)))))

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