139. Word Break

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.

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.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        int sLen = s.length();
        
        int[] pos = new int[sLen+1];
        Arrays.fill(pos, -1);
        pos[0] = 0;

        for(int i = 0; i < sLen; i++) {
            if(pos[i] != -1) {
                for(int j = i+1; j <= sLen; j++) {
                    String str = s.substring(i, j);
                    if(wordDict.contains(str)) {
                        pos[j] = i;
                    }
                }
            }
        }

        return pos[sLen] != -1;
    }
}

Solution 2

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        if(s.length() > 100) return false;
        for(String word: wordDict) {
            if(s.startsWith(word) && 
               (s.equals(word) || 
                   wordBreak(s.substring(word.length()), wordDict))) {
                return true;
            }
        }
        return false;
    }
}

Follow up