140. Word Break II

Description

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.

 

Solutions

Solution: Backtracking

• Time complexity: O(2ⁿ)
• Space complexity: O(2ⁿ)

 

JavaScript

/**
 * @param {string} s
 * @param {string[]} wordDict
 * @return {string[]}
 */
const wordBreak = function (s, wordDict) {
  const n = s.length;
  const wordDictSet = new Set(wordDict);
  const subStrMap = new Map();
  const result = [];
  const breakWord = (start, current) => {
    if (start === n) {
      result.push(current.trimEnd());
      return;
    }
    for (let index = start; index < n; index++) {
      const key = `${start},${index + 1}`;
      const memoSubStr = subStrMap.get(key);
      const subStr = memoSubStr ?? s.slice(start, index + 1);

      if (!memoSubStr) subStrMap.set(key, subStr);
      if (!wordDictSet.has(subStr)) continue;
      breakWord(index + 1, `${current}${subStr} `);
    }
  };

  breakWord(0, '');
  return result;
};