Word Break II

This page explains Java solution to problem Word Break II using Recursion with memoization.

Problem Statement

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.
• You may assume the dictionary does not contain duplicate words.

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

Input: s = "pineapplepenapple", wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
Output:
[
      "pine apple pen apple",
      "pineapple pen apple",
      "pine applepen apple"
]

Solution

package com.vc.hard;

import java.util.*;

class WordBreakIi {
    private HashMap<Integer, List<String>> memo = new HashMap<>();

    public List<String> wordBreak(String s, List<String> wordDict) {
        HashSet<String> words = new HashSet<>(wordDict);

        memo.put(s.length(), Collections.singletonList(""));
        return helper(s, 0, words);
    }

    private List<String> helper(String s, int from, HashSet<String> words) {
        if(memo.containsKey(from)) return memo.get(from);

        List<String> subRes = new ArrayList<>();
        for(int len = 1; from + len <= s.length(); len++) {
            String prefix = s.substring(from, from + len);
            if(words.contains(prefix)) {
                List<String> remaining = helper(s, from + len, words);
                for(String remainingStr: remaining) {
                    String output = remainingStr.length() > 0 ? prefix + " " + remainingStr : prefix;
                    subRes.add(output);
                }
            }
        }
        memo.put(from, subRes);
        return subRes;
    }
}

Time Complexity

O(N! + W) Where
N is total number of elements in an input String s.
W is total number of elements in an input array wordDict.

Space Complexity

O(W + 2^N) Where
N is total number of elements in an input String s.
W is total number of elements in an input array wordDict.