Word Break

Clarify the problem:

• The problem requires us to determine whether a given string can be segmented into a space-separated sequence of one or more dictionary words.
• We need to return a boolean value indicating whether such segmentation is possible.

Analyze the problem:

• Input: A string s and a list of dictionary words.
• Output: A boolean value indicating whether the string s can be segmented into dictionary words.
• Constraints:
  • The length of the string s is between 1 and 300.
  • The number of dictionary words is between 1 and 5,000.
  • The length of each word is between 1 and 20.
  • The characters in the string and dictionary words are lowercase English letters.

Design an algorithm:

• We can solve this problem using a dynamic programming approach.
• We create a boolean array dp of size equal to the length of the string s.
• Each element dp[i] represents whether a substring ending at index i (inclusive) can be segmented into dictionary words.
• We initialize dp[0] as True since an empty string can be segmented.
• We iterate over the characters of the string s and update the dp array.
• At each iteration, we check if there is a word in the dictionary that matches the substring from the current index to the previous indices where dp is True.
• If such a word exists, we mark dp[i] as True.
• Finally, we return dp[-1], which represents whether the entire string can be segmented into dictionary words.

Explain your approach:

• We will implement a dynamic programming solution to determine whether the string can be segmented into dictionary words.
• We initialize a boolean array dp of size equal to the length of the string.
• We set dp[0] as True since an empty string can be segmented.
• We iterate over the characters of the string.
• For each character at index i, we iterate from 0 to i (inclusive) to find a word in the dictionary that matches the substring.
• If such a word exists, and dp[j] is True, where j is the index before the current index i, we mark dp[i] as True.
• Finally, we return dp[-1].

Write clean and readable code:

python

def wordBreak(s, wordDict):
    n = len(s)
    dp = [False] * (n + 1)
    dp[0] = True

    for i in range(1, n + 1):
        for j in range(i):
            if dp[j] and s[j:i] in wordDict:
                dp[i] = True
                break

    return dp[-1]

Test your code:

• We can test the code with multiple test cases, including edge cases and corner cases, to ensure its correctness.
• For example:

  python

• # Example 1
  s = "leetcode"
  wordDict = ["leet", "code"]
  print(wordBreak(s, wordDict))
  # Output: True
  
  # Example 2
  s = "applepenapple"
  wordDict = ["apple", "pen"]
  print(wordBreak(s, wordDict))
  # Output: True
  
  # Example 3
  s = "catsandog"
  wordDict = ["cats", "dog", "sand", "and", "cat"]
  print(wordBreak(s, wordDict))
  # Output: False
  

Optimize if necessary:

• The solution has a time complexity of O(n^3), where n is the length of the input string.
• This is because we have two nested loops, each iterating up to the length of the string.
• The space complexity is O(n) as we use the dp array of size n+1.

Handle error cases:

• There are no specific error cases to handle for this problem.

Discuss complexity analysis:

• The time complexity of the solution is O(n^3) where n is the length of the input string.
• The space complexity is O(n) as we use the dp array of size n+1.
• The solution is efficient for the given constraints and provides the correct output.