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

(define/contract (word-break s wordDict)
  (-> string? (listof string?) boolean?)
  (let ((memo (make-hash)))
    (define (dp i)
      (cond
        [(= i (string-length s)) #t]
        [(hash-has-key? memo i) (hash-ref memo i)]
        [else
         (let loop ((words wordDict))
           (if (null? words)
               (begin (hash-set! memo i #f) #f)
               (let* ((word (car words))
                      (len (string-length word)))
                 (if (and (<= (+ i len) (string-length s))
                          (string=? (substring s i (+ i len)) word)
                          (dp (+ i len)))
                     (begin (hash-set! memo i #t) #t)
                     (loop (cdr words))))))]))
    (dp 0)))

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