Word Break
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
Explanation: No valid segmentation exists for "catsandog".
Examples
Example 1
Input: s = "leetcode", wordDict = ["leet","code"]
Output: true
Explanation: "leetcode" can be split as "leet" + "code", both of which are in the dictionary.
Example 2
Input: s = "applepenapple", wordDict = ["apple","pen"]
Output: true
Explanation: "applepenapple" can be split as "apple" + "pen" + "apple". Dictionary words can be reused.
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
Optimal Complexity
Time
O(n^2 * m)
Space
O(n)
Practice this problem with an AI interviewer
TechInView conducts a full voice mock interview — the AI asks clarifying questions, evaluates your approach, watches you code, and scores you on 5 dimensions. Just like a real FAANG interview.
Start a free interview