Problem description:

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 assume the dictionary does not contain duplicate words.

For example, given
s = “leetcode”,
dict = [“leet”, “code”].

Return true because “leetcode” can be segmented as “leet code”.

UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

Solution:

The idea for the solution is that a given string s can be divided into two substring s1 and s2. If s1 and s2 both fulfill the requirement, then complete the problem. Let’s create a dp array, which dp[i]= 1 means we are able to find match in dictionary.

As we just said, the original string can treat as substrings. Therefore, we can use a two pointer to check whether if there’s matches in a string.

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class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        # DP
        # subproblem answer lead to final one
        # if we can find x is s[i+1:] in wordDict after completed s[0:i]
        # then return true
        dp = [False] * (len(s)+1)
        dp[0] = True
        for i in range(1, len(s)+1):
            for j in range(i):
                if dp[j] and s[j:i] in wordDict:
                    dp[i] = True
                    break
        return dp[-1]
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class Solution {
public:
    //https://leetcode.com/problems/word-break/discuss/43814/C++-Dynamic-Programming-simple-and-fast-solution-(4ms)-with-optimization
    bool wordBreak(string s, vector<string>& wordDict) {
        
        if(wordDict.size()==0) return false;
        
        vector<bool> dp(s.size()+1,false);
        dp[0]=true;
        unordered_set<string> dict(wordDict.begin(), wordDict.end());
        
        for(int i=1;i<=s.size();i++)
        {
            for(int j=i-1;j>=0;j--)
            {
                if(dp[j])
                {
                    string word = s.substr(j,i-j);
                    if(dict.find(word)!= dict.end())
                    {
                        dp[i]=true;
                        break; //next i
                    }
                }
            }
        }
        
        return dp[s.size()];
    }
};

Second time:

similar idea from previous one.

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

Solution BFS

Modeled as a graph problem - every index is a vertex and every edge is a completed word
For example, the input string is “nightmare”, there are two ways to break it, “night mare” and “nightmare”. The graph would be

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0-->5-->9

|__ __ _^

The question is simply to check if there is a path from 0 to 9. The most efficient way is traversing the graph using BFS with the help of a queue and a hash set. The hash set is used to keep track of the visited nodes to avoid repeating the same work.

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class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        queue = deque()                                                                           
        visited = set()                                                                                     
        queue.append(0)
        visited.add(0)
        while len(queue) > 0:
            curr_index = queue.popleft()
            for i in range(curr_index, len(s)+1):
                if i in visited:
                    continue
                if s[curr_index:i] in wordDict:
                    if i == len(s):
                        return True
                    queue.append(i)
                    visited.add(i)
        return False