leetcode -- Word Break

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

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

Return true because "leetcode" can be segmented as "leet code".

[解题思路]

1.brute force will TLE.

just check every substring from index 0 to the end.

public boolean wordBreak(String s, Set<String> dict) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int len = s.length();
        boolean flag = false;
              
        for(int i = 1; i <= len; i++){
            String subStr = s.substring(0, i);
            if(dict.contains(subStr)){
                if(subStr.length() == s.length()){
                    return true;
                }   
                flag = wordBreak(s.substring(i), dict);
            }
        }
        return flag;
    }

2.DP

Reference the dicussion in leetcode.

Here we use seg(i, j) to demonstrate whether substring start from i and length is j is in dict?

base case:

when j = 0; seg(i, j) = false;

State transform equation:

seg(i, j) = true. if s.substring(i, j - 1) is in dict.

else seg(i, j) = seg(i, k) && seg(i + k, j - k);

public boolean wordBreak(String s, Set<String> dict) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(s == null || dict.size() <= 0){
            return false;
        }
        
        int length = s.length();
        // seg(i, j) means substring t start from i and length is j can be segmented into 
        // dictionary words
        boolean[][] seg = new boolean[length][length + 1];
        for(int len = 1; len <= length; len++){
            for(int i = 0; i < length; i++){
                String t = s.substring(i, i + len);
                if(dict.contains(t)){
                    seg[i][len] = true;
                    continue;
                }
                for(int k = 1; k < len; k++){
                    if(seg[i][k] && seg[i+k][len-k]){
                        seg[i][len] = true;
                        break;
                    }
                }
            }
        }
        return seg[0][length];
    }

这题貌似是一道面试题：http://thenoisychannel.com/2011/08/08/retiring-a-great-interview-problem/