Problem link:
http://oj.leetcode.com/problems/word-break/
We solve this problem using Dynamic Programming method. Let A[0..n-1] be a boolean array, where A[i]=True if and only if s[i..n-1] can be segmented into words. The recursive formula is:
A[i] = True, if s[i..n-1] is a word
A[i] = True, if there exists j > i such that s[i..j-1] is a word and A[j] == True
A[i] = False, otherwise
We fill the A-table from i=n to 0, and return A[0] to tell if s[0..n-1] can be segmented into words.
(Note: there is another way that A[i] means if s[0..i] can be segmented, then the recursive formula becomes a little different, we fill the table from i=0 to n, and return A[n-1])
The pseudo-code is as follows
WORD-BREAK(string s, dictionary d):
let A[0..n-1] be a new array of False
for i = n-1 to 0
if A[i..n-1] is a word in d
A[i] = True
else
for j = i+1 to n-1
if A[j] == True and s[i..j-1] is a word in d
A[i] = True
break
return A[0]
And 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 boolean
def wordBreak(self, s, dict):
"""
We solve this problem using DP
Define a boolean array A[0..n-1], where
A[i] = True, means s[i..n-1] can be segmented into words
------------------------------------
The recursive formula is:
A[i] = True, if there exists j>i (s[i..n-1] = s[i..j-1] + s[j..n-1])
such that s[i..j-1] is a word and A[j] = True
or A[i] = True, if A[i..n-1] is a word
------------------------------------
We fill A-table from i=n-1 to n
"""
n = len(s)
A = [False] * n
i = n-1
while i >= 0:
if s[i:n] in dict:
A[i] = True
else:
for j in xrange(i+1, n):
if A[j] and s[i:j] in dict:
A[i] = True
break
i -= 1
return A[0]