LeetCode-in-All

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

impl Solution {
    pub fn word_break(s: String, word_dict: Vec<String>) -> bool {
        let mut state = vec![None::<bool>; s.len() + 1];
        word_break_0(&s, &word_dict, &mut state)
    }
}

fn word_break_0(s: &str, word_dict: &Vec<String>, state: &mut Vec<Option<bool>>) -> bool {
    if s.is_empty() {
        return true;
    }

    if let Some(result) = state[s.len()] {
        return result;
    }

    let result = word_dict
        .iter()
        .filter_map(|word| s.strip_prefix(word))
        .any(|new_s| word_break_0(new_s, word_dict, state));

    state[s.len()] = Some(result);

    result
}

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