dict.add(“program”); }. current = current.children[index]; 41.2%: Medium: 140: Word Break II . Problem - 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. current = a.root; } System.out.println(” YES”); Trie() { I like this problem… it is so simple, but a nice exercise. Output: false Leetcode: Longest Substring Without Repeating Char... Leetcode: Rotate List (complete project codes) (C++), Leetcode 48: Binary Tree Inorder Traversal (C++). return true; TrieNode children[]; // There can be atmost 26 children (english alphabets) Here is a version using a stack instead of recursion (just for fun), however the complexity is O(n^2)… not acceptable. sbOne.delete(start,counter); This implementation looks neat. dict.add("code"); Appreciate it! map.put(s.substring(i), true); import java.util. return false; System.out.print(firstWord); return true; } TrieNode current = root; for(int i=0;i s.length()) If you want to ask a question about the solution. Apparetly, this algorithm is not correct. this.val = val; int start=0, end=str.length(), counter=1; while(counter <= end){ boolean result = this.wordBreak(newS, dict); Another solution , O(n^3), being n the length() of s. I assume that the set is a hashSet. break; Very short Python solution, also using trie: self.children = [None for i in range(ord(“z”) – ord(“a”) + 1)]. current = a.root; } Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Binary Tree 2.4. June 1, 2015 June 1, 2015 zn13621236 Leave a comment. if (start == n) fahsrouq created at: November 30, 2020 5:29 PM | No replies yet. Time Complexity : O(n) + O(m) } 4752 239 Add to List Share. current.children[index] = new TrieNode(input.charAt(i)); current = current.children[index]; boolean isRoot; TrieNode current = root; for(int i=0;i dict, int start){ The ascend order of the first item in MAP class: s... Leetcode:Two sum (9ms)(hash table)(Analysis & solu... Leetcode: Merge Intervals: Analysis and solution ... some tricks in using vector.size() in C++. Word Break II. 140. int[] pos = new int[s.length()+1]; }. } for (int i = 0; i < s.length(); i++) { For example, given s = "catsanddog", dict = ["cat", "cats", "and", "sand", "dog"], the solution is … boolean[] pos = new boolean[s.length()+1]; } int len = a.length(); Trie a = new Trie(); current.isLeaf = true; LeetCode Curated Algo 170 LeetCode Curated SQL 70 Top 100 Liked Questions Top Interview Questions ️ Top Amazon Questions Top Facebook Questions ⛽ Top Google Questions Ⓜ️ Top Microsoft Questions. } When you call dict.contains() in solution 3, I think below the surface the dictionary is looped through too. Word Break II - LeetCode. TrieNode current = a.root; if (s == null || s.length() == 0) char val; String subWord = s.substring(j + 1, j + a); One of the questions will be: Can we use the same dictionary word more than once? [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. initializeChildren(); } }, if(current.children[index] == null) /* package whatever; // don’t place package name! Input: s = "leetcode", wordDict = ["leet", "code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code". isLeaf = false; // Start from the beginning for the next character System.out.println(wordBreak(“leetcodesamsung”)); if (wordDict.contains(s.substring(i, j))) //Because we need initial state Set dict = new HashSet(); For example, given } dict = ["leet", "code"]. startIndex = endIndex; For INPUT: “leetcode”, [“leetcode”,”leet”,”code”]. arr.add(sb.substring(wordIndex, i)); 1 min read. pos[0]=0; children = new TrieNode[26]; Word Break 四种方法 中文解析 - Duration: 20:19. t[end] = true; Add Two Numbers (Medium) 3. if (wordBreak(“abcde”, dict)) { public boolean wordBreak(String s, Set dict) { if(words.contains(sbOne.substring(start,counter))){ System.out.println("Wordbreak (leetcode) = " + temp.wordBreak("leetcode", dict)); System.out.println("Wordbreak (lesscode) = " + temp.wordBreak("lesscode", dict)); Small correction of complexity in the 2nd case. Hard. } current = current.children[index]; dict.add("programcree"); int endIndex = i+1; Note: The same word in the dictionary may be reused multiple times in the segmentation. Example 2: Input: s = "applepenapple", wordDict = ["apple", "pen"] Output: true Explanation: Return true because " applepenapple " can be segmented as " apple pen apple " . // Word is not in the dictionary For example, given s = "helloworld", dict = ["world", "hello"]. for(String a: dict){ In Solution 2, if the size of the dictionary is very large, the time is bad. if(t[end]) continue; return true; if (isSame) } For example, given s = "leetcode", dict = ["leet", "code"]. 1 min read. int start = 'A'; for(int i=0; is.length){, if(dic.get(sb.substring(wordIndex,i) != null){ dict.add(“de”); The dynamic solution can tell us whether the string can be broken to words, but can not tell us what words the string is broken to. Java 7, substring ( ) in solution 2, if the size of the algorithm `` leet '' dict... 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