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The Word Break II problem in Python

The Word Break II problem is about constructing all possible sentences, with words that are all present in the dictionary, by adding spaces in a string. A word in the dictionary can be used multiple times as well.

For instance, if the dictionary contains the words “cookie”, “cookies”, “and”, “sand”, “cream” – and the string s is “cookiesandcream”-- then the valid sentences are “cookie sand cream” and “cookies and cream”.

Algorithm

Dynamic Programming is used to solve the word break II problem; the idea is to store and re-use the partial results.

Let result[j-1] contain all the valid sentences that can be formed with a substring of string s of length $j$. Then the valid sentences, for s and length $l$, can be found out by checking if the substring, s[j : l], is present in the dictionary and appending this word to all the sentences in result [j-1]. String s

Since “cream” is a valid dictionary word, append it to all the sentences that can be formed by “cookiesand”.

The animation below shows a few iterations of the algorithm in action.

1 of 22

Implementation

• i is set to j-1 and, in every iteration, is incremented by 1 to check all the substrings ending at the index j-1 (line 6).
• If a substring is a word in the dictionary, and the substring is not starting from index 0 (the start of s), then check if the substring starting from index 0 of s, that ends just before index i, forms a sentence. If it does, append the word to all the sentences in the result[i-1] (lines 16 - 24, line 34) .
• If a substring is a word in the dictionary and the substring is starting from index 0 (the start of s), then append the word to an empty list and append this list to the result (lines 26 - 27).
• Else, append an empty list to the result to indicate that a valid sentence could not be formed by the substring of length j starting at index 0 (lines 31 - 32).
• In the end, check if s is a valid dictionary word. If so, append it to the result (lines 35 - 36).
result = []
max_l = len(max(dictionary, key=len))
length = len(s) + 1
for j in range(1,length):
i = j - 1
flag = 0
ans = []
x = 0
# Letting setting x to j - max_l optimization,
# the code will work even if x is always set to 0
if j > max_l:
x = j - max_l
while(i >= x):
if s[i:j] in dictionary:
if i > 0 and result[(i - 1)]:
# appending the word to all the valid sentences
# formed by the substring ending at i-1
temp = list((map(lambda x : x + " "+ s[i:j],\
result[(i - 1)])))
for elem in temp:
ans.append(elem)
flag = 2
else:
flag = 1
result.append([s[i:j]])
i=i-1
# if the substring does not belong to the
# dictionary append an empty list to result
if flag == 0:
result.append([])
if flag == 2:
result.append(ans)
if s in dictionary:
result[len(s) - 1].append(s)
# Printing the result.
temp = ", result [{}]: "
for i in range(len(s)):
print("s:", s[:(i+1)], temp.format(i), result[i])
# If result[len(s)-1] is empty then the string cannot be
# broken down into valid strings

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