word break

139 Word Break

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.
Note:
* The same word in the dictionary may be reused multiple times in the segmentation.
* You may assume the dictionary does not contain duplicate words.

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




// dfs 
class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
      if(s == null || wordDict == null) return false;
      
      if(s.equals("")) return true;
      
      for(int i = 0; i < s.length(); i++){
        String substr = s.substring(0, i + 1);
        if(wordDict.contains(substr)){
          if(wordBreak(s.substring(i + 1), wordDict)){
            return true;
          }
          wordDict.remove(substr);
        }
      }
      return false;        
    }
}





// dp 

// Time complexity : O(n^2)
// Two loops are their to fill dp array.

//Space complexity : O(n). Length of pp array is n+1

 

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




dp builds up from bottom 

s= “”+ leetcode
f=[0     00000000]

Initalize 

f[0]   —————————breakable 
“”   breakable 



f[1]  —————————not breakable 
“”+ l 
“” Breakable , l is not in the dict
f[1] remains 0, which means false 



f[2]  —————————not breakable 
“”+ le
“” Breakable , le is not in the dict , 
“” + l is not breakable , f[1]


f[3]  —————————not breakable 
“”+ lee
“” Breakable  f[0], lee is not in the dict , 
“” + l is not breakable f[1] 
“”+ le is not brekable f[2]


f[4]  ————————— breakable 
“”+ leet

“” Breakable f[0] , leet is  in the dict 




f[5]  —————————not breakable 
“”+ leetc
“” Breakable , leetc is not in the dict 
“”+l, not breakable f[1] 
“”+ le is not brekable f[2]
“” + lee not breakable f[3]
“” + leet breakable f[4], c is not in the dict 



f[6]  —————————not breakable 
“”+ leetco
“” Breakable , leetco is not in the dict 
“”+l, not breakable f[1] 
“”+ le is not brekable f[2]
“” + lee not breakable f[3]
“” + leet breakable f[4], co is not in the dict 
“” + leetc not breakable f[5]


f[7]  —————————not breakable 
“”+ leetcod
“” Breakable , leetcod is not in the dict 
“”+l, not breakable f[1] 
“”+ le is not brekable f[2]
“” + lee not breakable f[3]
“” + leet breakable f[4], cod is not in the dict 
“” + leetc not breakable f[5]
“” + leetco not breakable f[6]


f[8]  ————————— breakable 
“”+ leetcode
“” breakable , leetcod is not in the dict 
“”+l, not breakable f[1] 
“”+ le is not brekable f[2]
“” + lee not breakable f[3]
“” + leet breakable f[4], code is  in the dict