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139. Word Break

Medium

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

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

Constraints:

• 1 <= s.length <= 300
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 20
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.

Solution

function wordBreak(s: string, wordDict: string[]): boolean {
    const dp: boolean[] = []
    for (let i = 0; i <= s.length; i++) dp.push(false)
    dp[s.length] = true
    for (let j = s.length - 1; j >= 0; j--) {
        for (const word of wordDict) {
            if (s.slice(j, j + word.length) === word && j + word.length <= s.length) {
                dp[j] = dp[j + word.length]
            }
            if (dp[j]) break
        }
    }
    return dp[0]
}

export { wordBreak }

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