What if you could instantly know if a long jumbled sentence is made of real words without trying every split?
Why Word Break Problem in DSA Python?
Imagine you have a long sentence without spaces, like "applepenapple", and you want to find out if it can be split into valid words from a dictionary, such as "apple" and "pen".
Trying every possible way to split the sentence manually is slow and confusing. You might miss some combinations or spend too much time checking all splits, especially if the sentence is long.
The Word Break Problem uses a smart way to remember which parts of the sentence can be split successfully. This avoids repeating the same checks and quickly tells if the sentence can be formed from dictionary words.
def can_break(s, word_dict): for i in range(1, len(s)): if s[:i] in word_dict and can_break(s[i:], word_dict): return True return s in word_dict
def can_break(s, word_dict): dp = [False] * (len(s) + 1) dp[0] = True for i in range(1, len(s) + 1): for j in range(i): if dp[j] and s[j:i] in word_dict: dp[i] = True break return dp[len(s)]
This concept enables fast and reliable checking if a sentence can be split into dictionary words, even for long sentences.
Spell checkers and text input tools use this to suggest spaces or correct words when users type without spaces or with typos.
Manual splitting is slow and error-prone.
Word Break uses memory to avoid repeated checks.
It quickly finds if a sentence can be formed from dictionary words.
Practice
Word Break Problem in algorithms?Solution
Step 1: Understand the problem statement
The Word Break Problem asks if a given string can be split into valid words from a dictionary.Step 2: Match the goal with options
To check if a string can be segmented into a sequence of dictionary words correctly describes this goal, while others describe unrelated tasks.Final Answer:
To check if a string can be segmented into a sequence of dictionary words -> Option AQuick Check:
Word Break Problem = Check segmentation [OK]
- Confusing Word Break with string reversal
- Thinking it counts vowels or finds longest word
- Assuming it modifies the string instead of checking segmentation
s[:i] is in a dictionary wordDict inside a loop?Solution
Step 1: Understand Python membership syntax
To check if an element is in a collection, useelement in collection.Step 2: Apply to substring and dictionary
We want to check if substrings[:i]is inwordDict, soif s[:i] in wordDict:is correct.Final Answer:
if s[:i] in wordDict: -> Option BQuick Check:
Membership test uses 'in' keyword [OK]
- Using 'contains' instead of 'in'
- Reversing order of operands in membership test
- Using equality '==' instead of membership
def wordBreak(s, wordDict):
dp = [False] * (len(s) + 1)
dp[0] = True
for i in range(1, len(s) + 1):
for j in range(i):
if dp[j] and s[j:i] in wordDict:
dp[i] = True
break
return dp[-1]
print(wordBreak('applepenapple', {'apple', 'pen'}))Solution
Step 1: Understand the dynamic programming approach
The code uses a list dp where dp[i] is true if s[:i] can be segmented into dictionary words.Step 2: Trace the input 'applepenapple'
'apple' and 'pen' are in the dictionary. The string can be segmented as 'apple' + 'pen' + 'apple', so dp[-1] becomes True.Final Answer:
True -> Option DQuick Check:
String segments into dictionary words = True [OK]
- Assuming false because of nested loops
- Confusing dp indices and substring slicing
- Ignoring the break statement after true found
def wordBreak(s, wordDict):
dp = [False] * len(s)
dp[0] = True
for i in range(1, len(s) + 1):
for j in range(i):
if dp[j] and s[j:i] in wordDict:
dp[i] = True
return dp[-1]Solution
Step 1: Check dp list initialization
dp should have length len(s) + 1 to include dp[0] for empty string and dp[len(s)] for full string.Step 2: Identify impact of incorrect dp size
With dp size len(s), dp[i] access when i == len(s) causes index error or incorrect logic.Final Answer:
dp list size is incorrect; should be len(s) + 1 -> Option AQuick Check:
dp size = len(s)+1 needed for indexing [OK]
- Initializing dp with length len(s) only
- Forgetting dp[0] represents empty string
- Not breaking loop after setting dp[i] true
s = 'catsanddog' and dictionary wordDict = {'cat', 'cats', 'and', 'sand', 'dog'}, which of the following lists contains all possible valid segmentations of s?Solution
Step 1: Identify valid dictionary words in the string
Words 'cat', 'cats', 'and', 'sand', 'dog' are in the dictionary.Step 2: Find segmentations matching dictionary words
Possible splits: 'cats' + 'and' + 'dog' and 'cat' + 'sand' + 'dog' are valid segmentations.Final Answer:
['cats and dog', 'cat sand dog'] -> Option CQuick Check:
All segments are dictionary words [OK]
- Ignoring valid splits and choosing whole string
- Mixing invalid word splits
- Confusing word boundaries
Ready to prove it?
Pass the test and this pattern moves unknown
