LeetCode-in-All

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

defmodule Solution do
  @spec min_distance(word1 :: String.t, word2 :: String.t) :: integer
  def min_distance(word1, word2) do
    for {ch1, i} <- [" " | String.codepoints(word1)] |> Enum.with_index(),
    {ch2, j} <- [" " | String.codepoints(word2)] |> Enum.with_index(),
    reduce: %{} do map ->
      cond do
        i == 0 -> j
        j == 0 -> i
        ch1 == ch2 -> Map.get(map, {i - 1, j - 1})
        true ->
          Map.get(map, {i - 1, j})
          |> min(Map.get(map, {i - 1, j - 1}))
          |> min(Map.get(map, {i, j - 1}))
          |> Kernel.+(1)
      end
      |> then(&(Map.put(map, {i, j}, &1)))
    end
    |> Map.get({String.length(word1), String.length(word2)})
  end
end

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