Word Break | LeetCode | Problem No. 139 | JAVA

Gowthaman Ravi
·Jun 18, 2021·

Subscribe to my newsletter and never miss my upcoming articles

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.

Approach: BFS

class Solution {
public static boolean wordBreak(String source, List<String> wordDict){
HashSet<String> set = new HashSet<>(wordDict);
int len = source.length();
boolean[] visited = new boolean[len];
while(!queue.isEmpty()){
int start = queue.remove();
for(int end = start +1; end<=len; end++){
if(visited[start])
continue;

if(set.contains(source.substring(start,end)) ){