LeetCode-in-All

139. Word Break

Medium

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

Example 1:

Input: s = “leetcode”, wordDict = [“leet”,”code”]

Output: true

Explanation: Return true because “leetcode” can be segmented as “leet code”.

Example 2:

Input: s = “applepenapple”, wordDict = [“apple”,”pen”]

Output: true

Explanation: Return true because “applepenapple” can be segmented as “apple pen apple”.

Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = “catsandog”, wordDict = [“cats”,”dog”,”sand”,”and”,”cat”]

Output: false

Constraints:

• 1 <= s.length <= 300
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 20
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.

Solution

defmodule Solution do
  @spec word_break(s :: String.t, word_dict :: [String.t]) :: boolean
  def word_break(s, word_dict) do
    Task.async(fn ->
      breakable?(s, 0, Enum.map(word_dict, &{&1, byte_size(&1)}))
    end)
    |> Task.await()
  end

  defp breakable?("", _idx, _words), do: true

  defp breakable?(s, idx, words) do
    memoized(idx, fn ->
      Enum.any?(words, fn {word, len} ->
        case s do
          <<^word::binary-size(len), rest::binary>> ->
            breakable?(rest, idx + len, words)

          _ ->
            false
        end
      end)
    end)
  end

  defp memoized(key, fun) do
    with nil <- Process.get(key) do
      fun.() |> tap(&Process.put(key, &1))
    end
  end
end

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