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.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.
• You may assume the dictionary does not contain duplicate words.

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

# @lc code=start
using LeetCode

# BFS
function word_break_bfs(s::String, word_dict::Vector{String})::Bool
    n, word_dict = length(s), sort!(word_dict; by=x -> length(x))
    valids, valid_pos = fill(false, n), [0]
    while !isempty(valid_pos)
        pos = popfirst!(valid_pos)
        for word in word_dict
            (new_pos = pos + length(word)) > n && break
            new_pos == n && s[(pos + 1):new_pos] == word && return true
            if !valids[new_pos] && s[(pos + 1):new_pos] == word
                valids[new_pos] = true
                push!(valid_pos, new_pos)
            end
        end
    end
    return false
end

# Dynamic Programming
function word_break(s::String, word_dict::Vector{String})::Bool
    lens = sort!(unique!(length.(word_dict)))
    dp = append!([true], fill(false, length(s)))
    for i in eachindex(dp)
        for len in lens
            i > len || break
            dp[i - len] && s[(i - len):(i - 1)] ∈ word_dict && (dp[i] = true)
        end
    end
    return last(dp)
end
# @lc code=end

word_break (generic function with 1 method)

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