Word Break LeetCode Solution – Queslers

Problem – Word Break

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.

Word Break LeetCode Solution in Java

public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
        
        boolean[] f = new boolean[s.length() + 1];
        
        f[0] = true;
        
        
        /* First DP
        for(int i = 1; i <= s.length(); i++){
            for(String str: dict){
                if(str.length() <= i){
                    if(f[i - str.length()]){
                        if(s.substring(i-str.length(), i).equals(str)){
                            f[i] = true;
                            break;
                        }
                    }
                }
            }
        }*/
        
        //Second DP
        for(int i=1; i <= s.length(); i++){
            for(int j=0; j < i; j++){
                if(f[j] && dict.contains(s.substring(j, i))){
                    f[i] = true;
                    break;
                }
            }
        }
        
        return f[s.length()];
    }
}

Word Break LeetCode Solution in C++

bool wordBreak(string s, unordered_set<string> &dict) {
        if(dict.size()==0) return false;
        
        vector<bool> dp(s.size()+1,false);
        dp[0]=true;
        
        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()];
    }

Word Break LeetCode Solution in Python

def word_break(s, words):
 	d = [False] * len(s)    
 	for i in range(len(s)):
 		for w in words:
 			if w == s[i-len(w)+1:i+1] and (d[i-len(w)] or i-len(w) == -1):
 				d[i] = True
 	return d[-1]

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