LeetCode - Word Break

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.

 

ChatGPT's Solution:

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        word_set = set(wordDict)  # Convert list to set to allow O(1) lookup
        n = len(s)
        dp = [False] * (n + 1)
        dp[0] = True  # Empty string is always "breakable"

        for i in range(1, n + 1):
            for j in range(i):
                if dp[j] and s[j:i] in word_set:
                    dp[i] = True
                    break  # No need to check further if s[0:i] is breakable
        
        return dp[n]

 

 

Here's the same implementation with detailed trace output that explains each step of the process:

def wordBreak(s: str, wordDict: list[str]) -> bool:
    word_set = set(wordDict)
    n = len(s)
    dp = [False] * (n + 1)
    dp[0] = True  # Base case: empty string is breakable

    print(f"String: '{s}'")
    print(f"Dictionary: {wordDict}")
    print(f"{'i':>2} {'s[j:i]':>10} {'dp[j]':>7} {'dp[i]':>7}  Explanation")

    for i in range(1, n + 1):
        for j in range(i):
            substring = s[j:i]
            if dp[j] and substring in word_set:
                dp[i] = True
                print(f"{i:>2} {substring:>10} {str(dp[j]):>7} {str(dp[i]):>7}  Found '{substring}' in dict, dp[{i}] = True")
                break
            else:
                print(f"{i:>2} {substring:>10} {str(dp[j]):>7} {str(dp[i]):>7}  ", end="")
                if not dp[j]:
                    print(f"dp[{j}] is False")
                elif substring not in word_set:
                    print(f"'{substring}' not in dict")
    print("\nFinal DP Table:", dp)
    return dp[n]

Example Run:

s = "applepenapple"
wordDict = ["apple", "pen"]
wordBreak(s, wordDict)

Sample Output:

String: 'applepenapple'
Dictionary: ['apple', 'pen']
 i   s[j:i]  dp[j]  dp[i]  Explanation
 1         a  False   False  dp[0] is False
 2        ap  False   False  dp[0] is False
 3       app  False   False  dp[0] is False
 4      appl  False   False  dp[0] is False
 5     apple   True    True  Found 'apple' in dict, dp[5] = True
...
Final DP Table: [True, False, False, False, False, True, ..., True]