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This is basically word break but they wanted the full treatment.
Start by clarifying the problem constraints and edge cases, then present a DP solution with a trie or hash set for O(1) word lookups, using prefix pruning to skip invalid starts. Discuss time/space complexity and how to extend to counting segmentations with modulo arithmetic.
Pro tip: Mention that early exits and pruning are crucial for large inputs, and that using a trie can reduce unnecessary substring checks, especially when the dictionary has many words with common prefixes.
Confirm that s and dict can be up to 100,000 entries, and that we need one valid segmentation or indication of impossibility. Ask about character set, case sensitivity, and whether empty strings are allowed.
Use a hash set for O(1) word lookups, or a trie to enable prefix pruning and reduce unnecessary substring checks. Discuss trade-offs: hash set is simpler but may check many substrings; trie can prune early.
Define dp[i] as whether s[0:i] can be segmented. Iterate i from 1 to n, and for each j < i, if dp[j] and s[j:i] in dict, set dp[i]=True and store parent pointer. Use early exit when a valid segmentation is found.
Time: O(n^2) worst-case with hash set, but with trie and pruning it can be closer to O(n * maxWordLength). Space: O(n) for DP and O(total characters) for trie. Mention that early exits can significantly reduce runtime in practice.
Modify DP to count ways: dp[i] = sum(dp[j] for j < i if s[j:i] in dict) mod 1e9+7. Use the same trie/hash set for lookups, and note that counting requires exploring all valid j, so early exits are not used.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.