【LeetCode OJ】Word Break II

Problem link:

http://oj.leetcode.com/problems/word-break-ii/

This problem is some extension of the word break problem, so the solution is based on the discussion in Word Break.

We also use DP to solve the problem. In this solution, A[i] is not a boolean any more, but a list of all possible value of j>i such that s[i..j-1] is a word and A[j]==True. The pseudo-code for computing the array is similar to that in Word Break with a few modifications

WORD-BREAK(string s, dictionary d):
  let A[0..n-1] be a new array
  for i = n-1 to 0
    if A[i..n-1] is a word in d
      A[i] = [n]
    else
      A[i] = [] // Empty set
      for j = i+1 to n-1
        if A[j] == True and s[i..j-1] is a word in d
            insert j to A[i]
  return A

The next step is to print all possible sentences with breaks. In another word, we need to find all valid sequences of breaks. Before doing this, lets review the meaning of A[i]. A[i] dentoes the next valid break after setting a break before s[i]. That is, we cannot set a break after we setting a break before s[i] if A[i] is []; otherwise A[i] is a list of all valid breaks after setting break before s[i].

A valid sequence of breaks should be in the form of (0, b1, ..., bm, n). We can solve all possible sequences by using BFS which starts from A[0] which contains valid values of the first break and find all valid paths (a path ends with "n") and print the sentences with the breaks of the path.

The path is at most of length |s|=n, and for each break there are at most |d| possible choices, so the BFS could terminate in O(|d|^|n|) time. The following code is the python solution accepted by oj.leetcode.

class Solution:
    # @param s, a string
    # @param dict, a set of string
    # @return a list of strings
    def wordBreak(self, s, dict):
        """
        We solve this problem using DP similar to the solution for word break I
        Let A[0..n-1] be an array, instead of being boolean, each A[i] is list of
        all possible j > i such that s[i..j-1] is a word and A[j] == True.
        """
        # Step 1: similar to word break I, 
        # but A[i] is a list instead of a boolean value
        n = len(s)
        A = [None] * n
        i = n-1
        while i >= 0:
            if s[i:n] in dict:
                A[i] = [n]  # A[i] contains "n" means s[i..n-1] is a word
            else:
                A[i] = []
            # Check al possible j break
            for j in xrange(i+1, n):
                if A[j] and s[i:j] in dict:
                    A[i].append(j)
            i -= 1

        # Step 2: find all possible sequences of breaks,
        # which equals to find all paths from A[0] and stop when the break is "n".
        # So it converts to BFS on a graph, with at most n steps.
        res = []  # possible sentences with break
        path_list = [[0]]  # initially, there is only one path containing the source node
        while path_list:
            new_list = []
            # For each path,
            # 1) If the path ends with break "n", then segment the string with breaks of the path
            # 2) otherwise, expand it with all possible breaks
            for path in path_list:
                if path[-1] == n:  # segment!
                    # Get words according to the breaks
                    temp = [ s[path[i]:path[i+1]] for i in xrange(len(path)-1) ]
                    # join words together
                    res.append(" ".join(temp))
                else:  # expand the path
                    for j in A[path[-1]]:
                        new_list.append(path+[j])
            path_list = new_list
        return res