139. Word Break

Problem Description

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.

Examples

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

def wordBreak(s: str, wordDict: List[str]) -> bool:
    n = len(s)
    dp = [False] * (n + 1)
    dp[0] = True
    word_set = set(wordDict)
    
    for i in range(1, n + 1):
        for j in range(i):
            if dp[j] and s[j:i] in word_set:
                dp[i] = True
                break
    
    return dp[n]

class Solution {
    public boolean wordBreak(String s, List wordDict) {
        int n = s.length();
        boolean[] dp = new boolean[n + 1];
        dp[0] = true;
        Set wordSet = new HashSet<>(wordDict);
        
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < i; j++) {
                if (dp[j] && wordSet.contains(s.substring(j, i))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        
        return dp[n];
    }
}

class Solution {
public:
    bool wordBreak(string s, vector& wordDict) {
        int n = s.length();
        vector dp(n + 1, false);
        dp[0] = true;
        unordered_set wordSet(wordDict.begin(), wordDict.end());
        
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < i; j++) {
                if (dp[j] && wordSet.count(s.substr(j, i - j))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        
        return dp[n];
    }
};

/**
 * @param {string} s
 * @param {string[]} wordDict
 * @return {boolean}
 */
var wordBreak = function(s, wordDict) {
    const n = s.length;
    const dp = new Array(n + 1).fill(false);
    dp[0] = true;
    const wordSet = new Set(wordDict);
    
    for (let i = 1; i <= n; i++) {
        for (let j = 0; j < i; j++) {
            if (dp[j] && wordSet.has(s.substring(j, i))) {
                dp[i] = true;
                break;
            }
        }
    }
    
    return dp[n];
};

public class Solution {
    public bool WordBreak(string s, IList wordDict) {
        int n = s.Length;
        bool[] dp = new bool[n + 1];
        dp[0] = true;
        HashSet wordSet = new HashSet(wordDict);
        
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < i; j++) {
                if (dp[j] && wordSet.Contains(s.Substring(j, i - j))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        
        return dp[n];
    }
}