LeetCode 139: Word Break

Problem Description

LeetCode Problem 139

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

Example 1:

Input: s = “leetcode”, wordDict = [“leet”,“code”]
Output: true
Explanation: Return true because “leetcode” can be segmented as “leet code”.

Example 2:

Input: s = “applepenapple”, wordDict = [“apple”,“pen”]
Output: true
Explanation: Return true because “applepenapple” can be segmented as “apple pen apple”.
Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = “catsandog”, wordDict = [“cats”,“dog”,“sand”,“and”,“cat”]
Output: false

Constraints:

• 1 <= s.length <= 300
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 20
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.

Difficulty: Medium

Tags: array, hash table, string, dynamic programming, trie, memoization

Rating: 95.51%

The Dynamic Programming Solution

Let’s break down the solution step by step:

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        # Convert wordDict to a set for O(1) lookups
        words = set(wordDict)
        
        # dp[i] represents if s[0:i] can be segmented using dictionary words
        dp = [False] * (len(s) + 1)
        
        # Empty string is always valid
        dp[0] = True
        
        # For each end position in the string
        for r in range(1, len(s) + 1):
            # Try all possible starting positions for the last word
            for l in range(r):
                # If we can form s[0:l] AND s[l:r] is in dictionary
                if dp[l] and s[l:r] in words:
                    dp[r] = True
                    break
        
        return dp[-1]

How the Algorithm Works

Let’s walk through an example with s = “leetcode” and wordDict = [“leet”, “code”]:

1. Initialize dp array: [True, False, False, False, False, False, False, False, False]
2. For r = 1 (“l”):
   • Check substrings: “l” → not in dictionary
   • dp[1] remains False
3. For r = 4 (“leet”):
   • Check “leet” → in dictionary and dp[0] is True
   • dp[4] becomes True
4. For r = 8 (“leetcode”):
   • When l = 4, we find that:
     • dp[4] is True (we can form “leet”)
     • “code” is in dictionary
   • Therefore, dp[8] becomes True

Complexity Analysis

• Time Complexity: $O(n²)$

  • We have a nested loop structure
  • Outer loop runs n times
  • Inner loop can run up to current position
  • String slicing and set lookup are O(1)

• Space Complexity: $O(n)$

  • We only need the dp array of size n+1
  • The word set takes additional space but is bounded by the dictionary size

Common Pitfalls and Optimization Tips

1. Converting to Set: Always convert the wordDict to a set for O(1) lookups. Using a list would result in O(n) lookups.
2. Breaking Early: The break statement is crucial for efficiency - once we find a valid segmentation for a position, we don’t need to check other possibilities.
3. Empty String Case: Handling the empty string case by setting dp[0] = True is essential for the algorithm to work correctly.

Additional Notes

• This solution is particularly efficient for shorter strings and moderate-sized dictionaries
• For very large strings or dictionaries, a trie-based solution might be more appropriate