Word Break

To solve this coding challenge, we need to determine if a given string

s

can be segmented into a space-separated sequence of one or more words found in a dictionary

wordDict

. This is known as the "Word Break Problem". We will use the dynamic programming approach to solve it.

Explanation:

Dynamic programming (DP) helps in breaking down problems into simpler subproblems. For this problem, we create a boolean DP array with a size of

len(s) + 1

, where each element represents whether the substring up to that index can be segmented into words from

wordDict

. The idea is:

1. Initialize the DP array with
   False
   values. Set
   dp[0]
   to
   True
   because an empty string can always be segmented.
2. Iterate through the string
   s
   with an outer loop.
3. Within the outer loop, use another inner loop to consider all possible substrings ending at the current position.
4. If a substring from the start of
   s
   to the current position can be found in
   wordDict
   and the substring before it can be segmented (checked using the DP array), mark the current position in the DP array as
   True
   .
5. Finally, the value of
   dp[len(s)]
   will tell us if the whole string can be segmented.
Here's the pseudocode with detailed comments:

                                            
# Function to determine if the string can be segmented
function wordBreak(string s, list wordDict):
# Create a DP array of size len(s) + 1, initialized to False
dp = new array of size (length of s + 1) with values False
# Base case: an empty string can always be segmented
dp[0] = True

# Outer loop to iterate over the string from 1 to len(s)
for i from 1 to length of s:
# Inner loop to consider all substrings ending at position i
for j from 0 to i - 1:
# Check if substring s[j:i] is in wordDict and if dp[j] is True
if dp[j] is True AND substring s[j:i] is in wordDict:
dp[i] = True
# No need to check further if we found a valid segmentation
break

# Return the value at the end of the DP array, which tells if the whole string can be segmented
return dp[length of s]

                                        

Step-by-Step Explanation:

6. Initialization :
• Create a DP array named
  dp
  with a size of
  len(s) + 1
  initialized to
  False
  .
• Set
  dp[0]
  to
  True
  , since an empty string can trivially be segmented.
7. Outer Loop :
• Iterate over each position
  i
  from
  1
  to
  len(s)
  . This index represents the current end position in the string that we are examining for potential segmentation.
8. Inner Loop :
• For each
  i
  , iterate over each position
  j
  from
  0
  to
  i-1
  . Here,
  j
  represents possible start positions for the substring ending at
  i
  .
9. Check Substring and DP Array :
• For each pair
  (i, j)
  , check if the substring
  s[j:i]
  is in the
  wordDict
  and if
  dp[j]
  is
  True
  .
• If both conditions are satisfied, set
  dp[i]
  to
  True
  because this means the substring
  s[0:i]
  can be segmented.
• Once
  dp[i]
  is set to
  True
  , break the inner loop as there is no need to test further substrings ending at
  i
  .
10. Final Result :
• After either exhausting or breaking out of the loops, return the value at
  dp[len(s)]
  . If it is
  True
  , it means the string can be segmented; otherwise, it cannot be.

This methodology ensures that we cover all possible ways to segment the string by leveraging efficient checks on whether previous segments were valid. The dynamic programming technique optimizes the problem by reusing previously computed results.