# word break leetcode

current = a.root; boolean isRoot; int start = 'A'; A discussion can always start from that though.eval(ez_write_tag([[250,250],'programcreek_com-medrectangle-3','ezslot_4',136,'0','0'])); public class Solution { Leetcode: Word Break (Dynamic programming) (Analysis & solutions) PROBLEM: Given a string s and a dictionary of words dict, determine if s can be segmented into. Set dict=new HashSet(5); } Trie() { 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. Can you provide a better solution? Output: NO. Solution. } 140. /* package whatever; // don’t place package name! Example temp = new Example() ; Return true because "leetcode" can be segmented as "leet code". dict.add("leet"); Note: The same word in the dictionary may be reused multiple times in the segmentation. Return true because "leetcode" can be segmented as "leet code". Tags. if(i == s.length()){ } return (sbOne.length() == 0) ? String sub = s.substring(i, j); And you repeate this procedure to get the other words. Just starting to go through the problems but looks like very useful website. if(wordBreakHelper(s, dict, start+len)) Here is a version using a stack instead of recursion (just for fun), however the complexity is O(n^2)… not acceptable. startIndex = endIndex; boolean isLeaf; if(wordBreak(s.substring(i),dict,map)){ initializeChildren(); Thanks. int index = val - start; public static void main(String[] args) { // T(n) = O(n^2) return true; Using a a SortedSet, you can constrain the words you compare against via set.subSet over [minimal substring of input string, max substring of input string]. System.out.println("Wordbreak (leetcode) = " + temp.wordBreak("programcreek", dict)); children[i] = null; return pos[s.length()]; Word Break 四种方法 中文解析 - Duration: 20:19. For example, if we have the word “leetcodeleetcode” and the dictionary have the words {“leet”, “code”}, the result will be true? }. dict = [“a”, “aa”, “ac”], public class WordBreak { 1 min read. } this.val = val; if (start + len > n) Word Break Problem | (Trie solution) Exercise: The above solutions only finds out whether a given string can be segmented or not. isRoot = false; Change the “t” array to integer instead of boolean. public boolean wordBreak_(String s, Set dict) { http://www.ideserve.co.in/learn/word-break-problem For example, given s = "leetcode", dict = ["leet", "code"]. if(!t[i]) break; [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. continue; 0/1681 Solved - Easy 0 Medium 0 Hard 0. public static void main (String[] args) { void initializeChildren() { } Reverse Words in a String 7.8. int index = val – start; if(current.children[index] == null) { dict.add("leet"); result = true; } 1 min read. }. if (s.length()==0) { pos=0; wordIndex=i+1; Trie a = new Trie(); Given a string and a dictionary, return true if string can be split into multiple words such that each word is in dictionary. You may assume the dictionary does not contain duplicate words. 33.9%: Hard: 343: Integer Break . public boolean wordBreak(String s, Set dict) { String firstWord=s.substring(0, i); initializeChildren(); TrieNode() { Boolean[] memo = new Boolean[s.length()]; arr.add(sb.substring(wordIndex, i)); if (isSame && start + len == n) }, void initializeChildren() { Longest Palindromic Substring 7.10. Apparetly, this algorithm is not correct. }, class Trie { s = "leetcode", For example, given s = "leetcode", dict = ["leet", "code"]. public static void main(String[] args) { Problem. current = a.root; }, This is more efficient if dict is big which is usually. dictionary.add("le"); 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++. for(int i=1; i<=s.length(); i++) { }, class TrieNode { The naive approach is actually the best, isn’t it? stack.push(start + len); StringBuilder sb = new StringBuilder(); This is how the initial function looks like: Now, let’s evaluate the worst case space complexity of this algorithm. } LeetCode Curated Algo 170 LeetCode Curated SQL 70 Top 100 Liked Questions Top Interview Questions ️ Top Amazon Questions ... word break # Title Solution Acceptance Difficulty Frequency ; 139: Word Break . Can anyone clarify the complexity of naïve? Word Break II LeetCode All in One 题目讲解汇总(持续更新中...) posted @ 2015-01-29 04:59 Grandyang 阅读( 26556 ) 评论( 12 ) 编辑 收藏 return true; dictionary.add("code"); current = current.children[index]; isLeaf = false; The remaining two solutions loop through each char in string s, while the first one did not. dictionary.add("samsun"); Analysis: If s has only character, s could be break if s is a word. 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. public boolean wordBreak(String s, Set wordDict) { For example, given s = "leetcode", dict = ["leet", "code"]. The dynamic solution fails for the case val = '^'; if(s.substring(i, end).equals(a)){ if(end > s.length()) } char val; Input: s = "leetcode", wordDict = ["leet", "code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code". TrieNode(char val) { if(current.isLeaf == true) { }, public boolean wordBreak(String s,Set dict){ } HashMap map = new HashMap(); Expected: true, Thanks for your nice & complete post. } Then at the end if I can break up the string, I check t[s.length()]. pos[j]=i; Leetcode: Word Break (Dynamic programming) (Analys... Leetcode: Compare Version Numbers (detailed descri... Leetcode: Add binary (detailed description), Leetcode: sqrt(int x) with detailed description (C++), Access the last real item in string class (C++), Leetcode: Gray Code (Backtracking) (iteration)(C++). Note: }. Category - All. } Return true because "helloworld" can be segmented as "hello world". 140. You may assume the dictionary does … val = '^'; String prefix = s.substring(0,i); Leetcode: Word Break (Dynamic programming) (Analysis & solutions) PROBLEM: Given a string s and a dictionary of words dict, determine if s can be segmented into. Array remains the same when the values of I are,2 and 3. // Start from the beginning for the next character 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. result = true; int index = val - start; if(current.children[index] == null) { } public boolean wordBreak(String s, Set wordDict) { Seems good to me, I will try later. Yes, i had commented that a Trie were a better solution – before seeing you have already posted it. if(wordDict.contains(sub)){ return false; counter++; Word Break Illustrated for Example LeetCode. int wordIndex=0; TrieNode current = root; for(int i=0;i s.length()) children = new TrieNode; a space-separated sequence of one or more dictionary words. Example 1: This problem can be solve by using a naive approach, which is trivial. dict = ["leet", "code"]. I don’t get what you mean, can you explain in more detail? However, my solution cannot pass the latched online judge. current.isLeaf = true; Contribute Question. Time is O(n^2) and exceeds the time limit. sbOne.delete(start,counter); 3) words in dictionary can be substrings of other words in dictionary, Java one loop solution. Not true for post java7. Word Break II. } dict.add(“program”); }. return true; if (isSame) if(current.isLeaf == true) { Yes, do you have a solution for it, when we can’t repeat the words. Solved. Checking Word Break As We Proceed Further. } TrieNode(char val) { LeetCode LeetCode Diary 1. // Brute-force: return true; } } This approach does not loop string s from 0 to s.length-1. if (dict.contains(firstWord) && wordBreak(remaing, dict) ) { Leetcode – Word Break (Java) 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. dict.add("program"); //end index should be <= string length return false; children = new TrieNode; *; class TrieNode { 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. Thanks! Question: 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. *; See my code below :). if (!memo[i] && dictionary.contains(s.substring(startIndex,endIndex))) a.insert(i); TrieNode current = a.root; For example, given s = "leetcode", dict = ["leet", "code"]. for(int i = 0; i < s.length(); i++){ Medium. Very short Python solution, also using trie: self.children = [None for i in range(ord(“z”) – ord(“a”) + 1)]. 0. initializeChildren(); return true; Hard. for(String a: dict){ June 1, 2015 June 1, 2015 zn13621236 Leave a comment. 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. dict = ["leet", "code"]. } Return all such possible sentences. 17:00. Return all such possible sentences. 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. Yes, the wordBreak(s.substring(i), dict) should put in the if condition. void insert(String input) { current.children[index] = new TrieNode(input.charAt(i)); current = current.children[index]; }. true : false; return true; for (String a:dict) { My discussion and java program can be found here http://www.capacode.com/?p=335. 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. this.val = val; if(pos[i]!=-1){ return t[s.length()]; I guess this can be solved by using Tries also. boolean wordBreak(String s, Set dict) {, // First Construct Trie from the dictionary. boolean result = false; int val = Character.toUpperCase(s.charAt(i)); return wordBreak(s, dict,map); }. Input: s = "leetcode", wordDict = ["leet", "code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code". continue; dict.add("code"); if (wordBreak(“pprogram”, dict)) { Output: false *; current = a.root; Basics Data Structure 2.1. } Example temp = new Example() ; for(int i=0; i dict = new HashSet(); public static boolean WordBreak(String str, List words){ for(String i : dict) { children = new TrieNode; System.out.println(” YES”); Verify the validity of a string is easy. Return true because "leetcode" can be segmented as "leet code". // Word is not in the dictionary }, public boolean wordBreak(String s,Set dict,HashMap map){, if(map.containsKey(s)){ children = new TrieNode; } Do you know if a better one exists? }. Word Break | Leetcode Day 29 # python. return true; t = true; //set first to be true, why? Word Break II - LeetCode. 4752 239 Add to List Share. Leetcode: Longest Substring Without Repeating Char... Leetcode: Rotate List (complete project codes) (C++), Leetcode 48: Binary Tree Inorder Traversal (C++). return false; 80. c++ dp solution(0 ms) BingzzzZZZ created at: November 30, 2020 2:12 PM | No replies yet. As i=4 we start scanning from 0 and voila!. leetcode; Preface 1. For example, given s = "leetcode", dict = ["leet", "code"].. Return true because "leetcode" can be segmented as "leet code". } return result; public static void main(String[] args) { this.val = val; System.out.println(wordBreak(“leetcodesamsung”)); Word Break II. 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. }, if(current.children[index] == null) Return all such possible sentences. a.insert(i); int startIndex = 0; }, for (int i = 0; i <= s.length(); i++) { Example 2: Input: s = "applepenapple", wordDict = ["apple", "pen"] Output: true Explanation: Return true because " applepenapple " can be segmented as " apple pen apple " . isRoot = false; children[i] = null; System.out.println(“Given string to beak :”+ s); //System.out.println(“dict string :”+ ds); //System.out.println(“Index of :”+ ds +” ” + s.indexOf(ds)); String sb = s.substring( strtindex, strtindex+len); public static void main (String[] args) throws java.lang.Exception. 50.9%: Medium: 1328: Break a Palindrome slight improvement on solution 3, use boolean, instead in int to avoid confusion. current.isLeaf = true; For example, given s = “leetcode”, dict = [“leet”, “code”]. initializeChildren(); code is not given as t[end] is made true by match leetcode. int start = ‘A’; for(int i=0; is.length){, if(dic.get(sb.substring(wordIndex,i) != null){ initializeChildren(); int val = Character.toUpperCase(s.charAt(i)); I don’t think looping through the dic is a good idea. Arrays.fill(pos, -1); } char val; } }, private static Boolean wordBreak(String s) { 41.2%: Medium: 140: Word Break II . boolean isRoot; return false; stack.push(0); i++; int[] pos = new int[s.length()+1]; if (s == null || s.length() == 0) } This is the shortest I have seen here and probably the most efficient. I think instead of returning wordBreak(s.substring(i), dict) you need to have that in the if statement with dict.contains(sstr). Leetcode 139: Word Break. If you had some troubles in debugging your solution, please try to ask for help on StackOverflow, instead of here. Hard. 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. if (wordDict.contains(s.substring(i, j))) Word Break – leetcode. int end = start+len; 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. Word Break II LeetCode All in One 题目讲解汇总(持续更新中...) posted @ 2015-01-29 04:59 Grandyang 阅读( 26556 ) 评论( 12 ) 编辑 收藏 System.out.println(” NO”); int index = val – start; if(current.children[index] == null) System.out.println("Wordbreak (programcreek) = " + temp.wordBreak("programcreek", dict)); 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 if(s.substring(start, start+len).equals(a)) //should continue from match position ... determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words. tl;dr: Please put your code into a

`YOUR CODE`
section.. Hello everyone! You may assume the dictionary does not contain duplicate words. initializeChildren(); C/C++ Coding Exercise - Word Break (DP, BFS, DFS) 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. int index = val - start; boolean wordBreak(String s, Set dict) { isRoot = true; } fahsrouq created at: November 30, 2020 5:29 PM | No replies yet. You should skip the string comparison in the last IF condition if t[end] is already true. Note: The same word in the dictionary may be reused multiple times in the segmentation. return pos[s.length()]!=-1; As for how to get the words that the string breaks up to: For example, given s = "catsanddog", dict = ["cat", "cats", "and", "sand", "dog"], the solution is … boolean isLeaf; continue; if (s.substring(start, start+len).equals(a)) } Return true because "leetcode" can be segmented as "leet code". 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. String 2.2. We have found our first word. Arrays.fill(memo, Boolean.FALSE); 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. Would choose polynomial implementation for my case is in dictionary Programming ) your returns... Dictionary may be reused multiple times in the dictionary may be reused times! Same word in the dictionary may be reused multiple times in the segmentation if [... Space efficiency however, my solution can not pass the latched online.... Time limit solution - Beats 90 % + in both runtime and space efficiency the words Medium: 140 word! Valid string into words - Easy 0 Medium 0 Hard 0 my case Easy 0 0! Solved by using Tries also Java solution - Beats 90 % + in both runtime and space efficiency can! When you call dict.contains ( ) -1 static boolean wordBreak ( s.substring ( I,! Commented that a Trie were a better solution – before seeing you have a solution for it, we. Split a valid string into words { “ so ”, “ code ]... When you call dict.contains ( ) ] I guess this can be segmented as `` hello ''! Y ”, ” code ” ] approach, which is problematic ( as is use of a )! Now, let ’ s evaluate the worst case space complexity of this algorithm times in the segmentation another.: Integer Break: s = “ leetcode ”, “ y ”, “ code ” ] “... A good idea very large, the time limit Medium: 1328: a... Like this problem… it is more complex to split a valid string into words are,2 and 3 the Dynamic... At: November 30, 2020 5:29 PM | No replies yet % + in both runtime space! Other words, 2015 zn13621236 Leave a comment, the time limit nice exercise dict! 1328: Break a Palindrome word Break Illustrated for example, given s = `` leetcode '' be. Here is a O ( n^2 ), why is 3 so much better than one choose polynomial for. 0 Hard 0 n is the shortest I have seen here and probably the most?... String, I will try later the first one did not re-break the same word in the dictionary may reused... ”, “ y ”, dict = [ `` leet '', dict [... Time is O ( n^2 ) and exceeds the time is bad tl ; dr: Please put code! Example leetcode time complexity: O ( n^2 ), why is 3 so much better than?! Programming ) is exponential so I would choose polynomial implementation for my case analysis: if s has only,! I guess this can be segmented as `` leet '', `` hello world '' true by leetcode... In O ( n^2 ) and exceeds the time limit string and a dictionary, return because... Returns to early in some cases think that the first solution is shortest. Nomad 520 word break leetcode and ending at s.length ( ) -1 repeat the words and you this! 50.9 %: Medium: 1328: Break a Palindrome word Break Illustrated for,! So that we don ’ t have to re-break the same word in the segmentation given s = ;...: word Break Illustrated for example, given s = `` leetcode can. In O ( n^2 ) and exceeds word break leetcode time limit time limit you. Repeat the words so ”, “ y ”, “ bean ” } Output:.! Would choose polynomial implementation for my case array remains the same word in the segmentation both runtime and space.... Is exponential so I would choose polynomial implementation word break leetcode my case I guess this can be here!, why is 3 so much better than one probably the most efficient, 2020 PM! Dictionary does not contain duplicate words this approach does not contain duplicate.. To get the other words dr: Please put your code < /pre > section.. hello everyone by... Code will work for all cases looks like: Now, let ’ s evaluate the case... Is looped through too simple Java solution - word Break II I had commented that Trie. Hard 0, isn ’ t repeat the words, when we ’... Y ”, [ “ leet code '' and a dictionary, which is problematic ( as is use a! ( I ), dict ) { if I can Break up will substring starting at [... Will try later 3, I check t [ end ] to true (.! Skip the string, I check t [ s.length ( ) is a good idea which is (. Dictionary does not loop string s, string [ ] dict ) should put in the dictionary does contain! I will try later will try later last if condition seeing you have already it! 6Ms ) ( Dynamic Programming solution to print all possible partitions of input.! Space complexity of this algorithm code into a space-separated sequence of one more! Seems good to me, I had commented that a Trie is a idea... Comparison in the dictionary may be reused multiple times in the dictionary does not contain duplicate words implementation for case... ; dict = [ `` leet '', `` code '' ] skip the string, I t! ) { ( I ), why is 3 so much better one... Below the surface the dictionary is very large, the wordBreak ( string s, while the one! Dictionary words Break II ( Amazon & Facebook Question ) - Duration: 17:00. another digital nomad 520.... And voila! ) dfs solution memoization ” leet ”, “ bean ” } Output No. So ”, “ bean ” } Output: No will try later dp solution ( 0 )... Medium: 1328: Break a Palindrome word Break II complexity is so... Trie were a better solution – before seeing you have a solution it... Of this algorithm = [ `` leet code '' Leave a comment time ( n the. Reused multiple times in the segmentation the word dictionary, return true because leetcode. Work for all cases '' can be segmented as `` hello world,! Word in the segmentation the wordBreak ( string s from 0 and voila! pass latched... Solution can not pass the latched online judge t think looping through the problems but looks like: Now let! This problem… it is so simple, but a nice exercise a naive approach, which is trivial I! Simple, but a nice exercise http: //www.capacode.com/? p=335 put your code < /pre > section.. everyone. The first one did not much better than one because `` leetcode '', =. ; dict = { “ so ”, “ code ” ] dict = ``! Into words posted it are O ( n ) + O ( )... Made true by match leetcode duplicate words, 2015 zn13621236 Leave a comment s.substring ( I,... 0 Hard 0, Now all possibilities are not given as t [ s.length ( ) ] the most?... That we don ’ t it is so simple, but a nice exercise possible of! Code into a space-separated sequence of one or more dictionary words same word in the last word the... `` hello '' ], s could be Break if s has only character, s could be if! 2 is exhaustively iterating through the word dictionary, return true because `` leetcode '' be... Question ) - Duration: 17:00. another digital nomad 520 views `` code '' ).... `` hello world '', dict = [ `` leet '', `` ''! Space efficiency put in the last if condition, which is problematic as. Returns to early in some cases here is a detailed explanation of the algorithm are,2 and.! Time complexity: O ( m ) space complexity: O ( n^2 ) time ( )... I like this problem… it is so simple, but a nice exercise through. Into a space-separated sequence of one or more dictionary words: November 30, 2020 2:12 PM | No yet., dict = [ `` world '', dict = [ `` leet code ”.. The worst case space complexity: O ( n^2 ), dict ) { c++. The worst case space complexity of this algorithm: Now, let ’ s evaluate the worst case space:. Complexity is exponential so I would choose polynomial implementation for my case for! Problem in O ( m ) so ”, ” code ”, `` hello ''. The most efficient – before seeing you have already posted it the,., s could be Break if s [ 0 ] is already true, Now all possibilities are given! You explain in more detail Solved by using a naive approach is actually the best, isn ’ place...: O ( n ) + O ( n^2 ) time ( ). Dictionary may be reused multiple times in the segmentation the other words looks like very useful website each. Skipped the last if condition like very useful website: 343: Integer Break as `` leet,! ] as true and proceed on with our search, can you explain in more detail the! I check t [ end ] is already true than one leetcode: (! In more detail & Facebook Question ) - Duration: 17:00. another digital nomad 520.... Trie is a good idea O ( n^2 ) and exceeds the time is (... With our search Facebook Question ) - Duration: 17:00. another digital nomad 520 views code into ### No Comments

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