140. Word Break II

140. Word Break II

Hard

Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order.

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

Example 1:

• Input: s = "catsanddog", wordDict = ["cat","cats","and","sand","dog"]
• Output: ["cats and dog","cat sand dog"]

Example 2:

• Input: s = "pineapplepenapple", wordDict = ["apple","pen","applepen","pine","pineapple"]
• Output: ["pine apple pen apple","pineapple pen apple","pine applepen apple"]
• Explanation: Note that you are allowed to reuse a dictionary word.

Example 3:

• Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
• Output: []

Constraints:

• 1 <= s.length <= 20
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 10
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.
• Input is generated in a way that the length of the answer doesn't exceed 105.

Solution:

class Solution { private $map = array(); /** * @param String $s * @param String[] $wordDict * @return String[] */ function wordBreak($s, $wordDict) { if(array_key_exists($s, $this->map)) { return $this->map[$s]; } $result = array(); if(strlen($s) == 0){ $result[] = ""; $this->map[""] = $result; return $result; } foreach($wordDict as $word) { if(strpos($s, $word) === 0){ $subWords = $this->wordBreak(substr($s, strlen($word)), $wordDict); foreach($subWords as $subWord) { $result[] = $word . (strlen($subWord) > 0 ? " " : "") . $subWord; } } } $this->map[$s] = $result; return $result; } } 

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