Скачать или смотреть 139 word break daily leetcode challenge august c java

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okay, let's dive into the leetcode "word break" problem. i'll provide a comprehensive guide with explanations, code examples in java, and considerations for the daily august leetcode context.

*problem statement (leetcode 139: word break)*

given a non-empty string `s` and a dictionary `worddict` containing a list of non-empty words, determine if `s` can be segmented into a space-separated sequence of one or more dictionary words. you may reuse words in the dictionary.

*example:*



*understanding the approach*

the core idea is to use dynamic programming (dp). we'll build a boolean array `dp` where `dp[i]` represents whether the substring `s[0...i-1]` can be segmented into words from the dictionary.

*algorithm*

1. *initialization:* `dp[0] = true` (an empty string can always be segmented).
2. *iteration:* iterate through the string `s` from index 1 to `s.length()`. for each index `i`, iterate through all possible starting points `j` (from 0 to `i-1`).
3. *check segmentation:* if `dp[j]` is `true` (meaning the substring `s[0...j-1]` can be segmented) and the substring `s[j...i-1]` is present in the `worddict`, then `dp[i] = true`. break the inner loop since we found a valid segmentation.
4. *result:* `dp[s.length()]` will hold the final answer (whether the entire string `s` can be segmented).

*java code*



*explanation:*

*`wordset`:* we convert the `worddict` to a `hashset` for faster `contains()` lookups (o(1) on average for a `hashset` vs. o(n) for a `list`).
*`dp` array:* `dp[i]` is `true` if the substring `s[0...i-1]` can be segmented using words from the dictionary.
*outer loop (i):* iterates from 1 to `s.length()` to consider all possible substring lengths from the beginning of `s`.
*inner loop (j):* iterates from 0 to `i-1` to try all possible starting positions `j` for a word ending at index `i-1`.
*`dp[j] && wordset.contains(s.substring(j, i))`:* this is the crucial check. if `dp[j]` is `tru ...

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