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”.

O(n^2) or O(n^3) time O(n) space DP solution

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
public class Solution {
    public boolean wordBreak(String s, Set<String> wordDict) {
        if (s == null || wordDict.size() == 0)
            return false;
        boolean[] breaks = new boolean[s.length() + 1];
        breaks[0] = true; // empty string should be true
        
        for (int i = 1; i <= s.length(); i++) {
            for (int j = 0; j < i; j++) {
                if (breaks[j] && wordDict.contains(s.substring(j, i))) {
                    breaks[i] = true; // we only care if this position can be broken, but don't care about how it should be break
                    break;
                }
            }
        }
        return breaks[s.length()];
    }
}

BFS Queue solution from Leetcode

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
public boolean wordBreak(String s, Set<String> dict) {
    if (dict.contains(s)) return true;
    Queue<Integer> queue = new LinkedList<Integer>();
    queue.offer(0);
    // use a set to record checked index to avoid repeated work.
    // This is the key to reduce the running time to O(N^2).
    Set<Integer> visited = new HashSet<Integer>();
    visited.add(0);
    while (!queue.isEmpty()) {
        int curIdx = queue.poll();
        for (int i = curIdx+1; i <= s.length(); i++) {
            if (visited.contains(i)) continue;
            if (dict.contains(s.substring(curIdx, i))) {
                if (i == s.length()) return true;
                queue.offer(i);
                visited.add(i);
            }
        }
    }
    return false;
}