Word Break
Given a string `s` and a dictionary `wordDict`, determine if `s` can be segmented into a space-separated sequence of one or more dictionary words.
Why Interviewers Ask This
IBM evaluates this question to assess a candidate's mastery of dynamic programming and recursive thinking. They specifically look for the ability to break complex string segmentation problems into overlapping subproblems, optimize solutions from exponential time complexity to polynomial, and handle edge cases like empty strings or non-existent words without crashing.
How to Answer This Question
1. Clarify constraints: Ask about string length limits and dictionary size to determine if O(N^2) is acceptable. 2. Define the state: Explicitly state that dp[i] represents whether the substring s[0...i-1] can be segmented. 3. Establish transitions: Explain that dp[i] is true if there exists any j < i where dp[j] is true AND s[j...i-1] exists in the dictionary. 4. Optimize space: Discuss using a hash set for O(1) word lookups instead of linear search. 5. Walk through an example: Trace 'leetcode' with ['leet', 'code'] step-by-step on a whiteboard to demonstrate logical flow before coding. This structured approach mirrors IBM's focus on methodical problem decomposition.
Key Points to Cover
- Explicitly define the DP state transition formula before writing code
- Demonstrate knowledge of converting naive recursion to memoization or iterative DP
- Utilize a Hash Set for constant-time dictionary lookups to optimize performance
- Handle base cases like empty strings or single character mismatches gracefully
- Explain the trade-off between time complexity O(n^2) and space complexity O(n)
Sample Answer
To solve the Word Break problem efficiently, I would use Dynamic Programming because brute force recursion leads to redundant calculations. First, I initialize a boolean array called dp of size n plus one, where dp[0] is true since an empty string is always valid. Next, I iterate through each index i from 1 to n. For every position i, I check all previous indices j from 0 to i-1. If dp[j] is true and the substring s[j...i] exists in our dictionary, then we know s[0...i] is also segmentable, so we set dp[i] to true and break the inner loop. Using a HashSet for the dictionary ensures O(1) average time complexity for lookups. For example, with s = 'applepenapple' and dict = ['apple', 'pen'], at i=5, we find 'apple' is valid. Then at i=8, we check back to j=5; since dp[5] is true and 'pen' is in the dict, dp[8] becomes true. Finally, we return dp[n]. This approach reduces time complexity to O(n^2) and space to O(n), which is optimal for standard interview constraints.
Common Mistakes to Avoid
- Attempting a pure recursive solution without memoization, leading to Time Limit Exceeded errors
- Failing to initialize dp[0] as true, which causes the algorithm to miss valid segments starting at index zero
- Using a list or array for the dictionary lookup instead of a Hash Set, resulting in unnecessary O(k) overhead per check
- Not considering edge cases where the input string contains characters not present in any dictionary word
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