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139. Word Break

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.

Solutions (Python)

1. Dynamic Programming

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        dp = [True] + [False] * len(s)

        for i in range(1, len(s) + 1):
            for j in range(i):
                if dp[j] and s[j:i] in wordDict:
                    dp[i] = True
                    break

        return dp[-1]

Solutions (Ruby)

1. Dynamic Programming

# @param {String} s
# @param {String[]} word_dict
# @return {Boolean}
def word_break(s, word_dict)
  dp = [true] + [false] * s.size

  (1..s.size).each do |i|
    (0...i).each do |j|
      if dp[j] && word_dict.include?(s[j...i])
        dp[i] = true
        break
      end
    end
  end

  dp[s.size]
end