Classic dp problem once you see it, but my first instinct was to go recursive without memoization and I had to backtrack.
Start by clarifying the problem constraints (e.g., can words be reused, case sensitivity, empty string). Then propose a dynamic programming solution where dp[i] indicates if the prefix of length i can be segmented, and for each i, check all dictionary words that end at i. Optimize with a trie or BFS to avoid redundant checks.
Pro tip: After presenting the DP solution, mention that you can optimize space to O(n) and time to O(n * maxWordLength) by using a trie or by iterating over word lengths present in the dictionary. Also, discuss trade-offs between DP and BFS/DFS with memoization.
Ask about input size, character set, whether words can be reused, and if the dictionary is static. Handle edge cases like empty string, empty dictionary, and very long strings.
Define dp[i] as whether the substring s[0..i-1] can be segmented. Initialize dp[0] = true. For each i from 1 to n, dp[i] = true if there exists j < i such that dp[j] is true and s[j..i-1] is in the dictionary.
Instead of checking all j, iterate over dictionary words and check if they match a suffix ending at i. Alternatively, use a trie to efficiently find all valid words ending at i, or use BFS with memoization to avoid redundant subproblems.
Naive DP is O(n^2 * L) where L is average word length. With a trie or by limiting to max word length, it becomes O(n * maxWordLength). Space is O(n). Compare with BFS/DFS approaches.
Walk through examples like 'leetcode' with ['leet','code'] and 'applepenapple' with ['apple','pen']. Also test cases where segmentation is impossible, e.g., 'catsandog' with ['cats','dog','sand','and','cat'].
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