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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