s.length()) Note: The same word in the dictionary may be reused multiple times in the segmentation. a space-separated sequence of one or more dictionary words. } June 1, 2015 June 1, 2015 zn13621236 Leave a comment. }. if (wordBreak(“abcde”, dict)) { children = new TrieNode; For example, given Word Break I. 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 TrieNode current = a.root; } 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. } import java.util.Set; public static boolean wordBreak(String s,Set dict) { } As i=4 we start scanning from 0 and voila!. 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. return true; if (isSame) dict.add(“code”); Change the “t” array to integer instead of boolean. if(i == s.length()){ Solution. initializeChildren(); counter=1; Seems good to me, I will try later. return wordBreakHelper(s, dict, 0); int start = 'A'; for(int i=0; i wordDict) { This problem can be solve by using a naive approach, which is trivial. int start = stack.pop(); 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; counter++; For example, given s = “leetcode”, dict = [“leet”, “code”]. TrieNode() { while (!stack.empty()) { dict.add("programcree"); Return all … I like this problem… it is so simple, but a nice exercise. for(String a: dict){ THe last word in the break up will substring starting at t[s.length()] and ending at s.length()-1. t[end] = true; Status. result = false; }, public static void main(String[] args) { import java.lang. return true; initializeChildren(); } }else{ I don’t think looping through the dic is a good idea. } Arrays.fill(memo, Boolean.FALSE); }, class TrieNode { int start = 'A'; 2) you can use a word for dictionary multiple times }. Return all such possible sentences. if(dict.contains(prefix)){ isRoot = false; } Why do I think that the first solution is the most efficient? 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. Substring with Concatenation of All Words, Leetcode: Triangle (6ms)(Dynamic programming). isRoot = false; // Brute-force: Reverse Words in a String 7.8. }. [leetcode]139. TrieNode current = a.root; char val; System.out.println(” NO”); Arrays.fill(pos, -1); children[i] = null; dict.add(“programcree”); Return true because "leetcode" can be segmented as "leet code". } Otherwise your function returns to early in some cases. 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. true : false; *; if (wordDict.contains(s.substring(i, j))) Space Complexity : O(m). public boolean wordBreak_(String s, Set dict) { boolean isLeaf; }, private static void initializeDictionary() { } return false; return true; for (String a:dict) { dict = ["leet", "code"]. isRoot = false; 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. Word Break II. if(s.substring(start, start+len).equals(a)) Return all such possible sentences. Leetcode 139: Word Break. dict.add("leet"); return true; } code is not given as t[end] is made true by match leetcode. Word Break II (Hard) 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. } Valid Palindrome 7.9. int start = ‘A’; int start = ‘A’; for(int i=0; i n) int len = a.length(); Word Break. dict.add(“de”); // Word is not in the dictionary return map.get(s); Return true because “leetcode” can be segmented as “leet code”. int len = a.length(); return true; wordIndex=i+1; } int val = Character.toUpperCase(input.charAt(i)); } Apparetly, this algorithm is not correct. continue; val = ‘^’; Checking Word Break As We Proceed Further. String sub = s.substring(i, j); You may assume the dictionary does not contain duplicate words. But the complexity is exponential so I would choose polynomial implementation for my case. String prefix = s.substring(0,i); public boolean wordBreak(String s, Set dict) {, if (dict.contains(s.substring(0, i + 1))) {. isRoot = true; Note: a.insert(i); Binary Tree 2.4. boolean isRoot; 80. c++ dp solution(0 ms) BingzzzZZZ created at: November 30, 2020 2:12 PM | No replies yet. This approach does not loop string s from 0 to s.length-1. dict.add(“leet”); 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.add("code"); sbOne.delete(start,counter); } else { children = new TrieNode; 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. In Solution 2, if the size of the dictionary is very large, the time is bad. } static Set dictionary = new HashSet(); void insert(String input) { }. 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++. Description. char val; 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 II (Java) Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. Thanks for these solutions. import java.util.HashSet; dict.add("program"); tl;dr: Please put your code into a
`YOUR CODE`
section.. Hello everyone! For example, given s = "leetcode", dict = ["leet", "code"].. Return true because "leetcode" can be segmented as "leet code". String 2.2. Todo. } } result = true; Analysis: If s has only character, s could be break if s is a word. }, public boolean wordBreak(String s,Set dict){ 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. public boolean wordBreakHelper(String s, Set dict, int start){ *; dict.add("creek"); System.out.println("Wordbreak (programcreek) = " + temp.wordBreak("programcreek", dict)); Word Break | Leetcode Day 29 # python. 0. public boolean wordBreak(String s, Set dict) { Set dict = new HashSet(); if(wordBreak(s.substring(i),dict,map)){ int start = stack.pop(); current = a.root; import java.util. Return true because "leetcode" can be segmented as "leet code". }. System.out.println(“Wordbreak (programcreek) = ” + temp.wordBreak(“programcreek”, dict)); if(wordDict.contains(sub)){ Simple Java Solution - Beats 90%+ in both runtime and space efficiency. The brute force solution seems to be wrong? } Word Break II - LeetCode. for(int i = 0; i < s.length(); i++){ } current.children[index] = new TrieNode(input.charAt(i)); int val = Character.toUpperCase(s.charAt(i)); For example, if we have the word “leetcodeleetcode” and the dictionary have the words {“leet”, “code”}, the result will be true? }. 17:00. }. } map.put(s.substring(i), true); Can you provide a better solution? } import java.io. Tags. slight improvement on solution 3, use boolean, instead in int to avoid confusion. } return true; if (s == null || s.length() == 0) System.out.println(“Wordbreak (leetcode) = ” + temp.wordBreak(“programcreek”, dict)); char val; dictionary.add("et"); Expected: true, Thanks for your nice & complete post. }, class Trie { Solved. dictionary.add("samsun"); t = true; //set first to be true, why? It’s mentioned that “Time: O(string length * dict size)” but you also run equals (and substring is not constant for Java > 1.6) for every word in dictionary so it’s more like O(string length * dict size * length of the longest word in dict). current = current.children[index]; fahsrouq created at: November 30, 2020 5:29 PM | No replies yet. Word Break 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. continue; if (s.substring(start, start+len).equals(a)) Hard. ... determine if s can be segmented into a space-separated sequence of one or more dictionary words. if (wordBreak(“pprogram”, dict)) { String subWord = s.substring(j + 1, j + a); One of the questions will be: Can we use the same dictionary word more than once? Then at the end if I can break up the string, I check t[s.length()]. Here is a version using a stack instead of recursion (just for fun), however the complexity is O(n^2)… not acceptable. 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. Solution 2 is exhaustively iterating through the word dictionary, which is problematic (as is use of a HashSet). You may assume the dictionary does not contain duplicate words. Here is a version using a stack instead of recursion (just for fun), however the complexity is O(n^2)… not acceptable. } For example, given s = "catsanddog", dict = ["cat", "cats", "and", "sand", "dog"], the solution is … if(s.substring(i, end).equals(a)){ int val = Character.toUpperCase(s.charAt(i)); for(int j=i+1; j<=s.length(); j++){ if (!memo[i] && dictionary.contains(s.substring(startIndex,endIndex))) Word Break Illustrated for Example LeetCode. Output: NO. Very short Python solution, also using trie: self.children = [None for i in range(ord(“z”) – ord(“a”) + 1)]. TrieNode current = root; for(int i=0;i> Interview March 8, 2014 Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. a.insert(i); TrieNode current = a.root; int i=0; 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.You may assume the dictionary does not contain duplicate words. Because you skipped the last IF condition if t[end] is already true, now all possibilities are not given. int start = 'A'; // Word is not in the dictionary 0. A Trie is a better solution than DP for this problem. dict.add(“program”); TrieNode(char val) { current = current.children[index]; Return true because "leetcode" can be segmented as "leet code". I like this problem… it is so simple, but a nice exercise. int index = val - start; if(current.children[index] == null) { 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. LeetCode 140. dict.add(“creek”); int index = val – start; if(current.children[index] == null) { We have found our first word. }, TrieNode(char val) { void initializeChildren() { int endIndex = i+1; For example, given s = "leetcode", dict = ["leet", "code"]. if (isSame && start + len == n) } } 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. val = '^'; }. a space-separated sequence of one or more dictionary words. boolean[] t = new boolean[s.length()+1]; Thanks. Hard. You may assume the dictionary does not contain duplicate words. Example temp = new Example() ; Longest Palindromic Substring 7.10. } public boolean wordBreak(String s, Set wordDict) { This is the shortest I have seen here and probably the most efficient. } Word Break II [Leetcode] Given a string s and a dictionary of words dict , add spaces in s to construct a sentence where each word is a valid dictionary word. current = a.root; Category - All. return false; Given a string and a dictionary, return true if string can be split into multiple words such that each word is in dictionary. } Let me know if the following code will work for all cases. System.out.println("Wordbreak (programcreek) = " + temp.wordBreak("programcreek", dict)); Return true because "leetcode" can be segmented as "leet code". return true; initializeChildren(); 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. isLeaf = false; // First Construct Trie from the dictionary. for (int i = 0; i < s.length(); i++) { if Approach 1 and Approach 3 both are O(N^2), why is 3 so much better than one? 1. 4752 239 Add to List Share. 68. } Example temp = new Example() ; public static void main(String[] args) { This implementation looks neat. } 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. It is more complex to split a valid string into words. Linked List 2.3. Word Break 四种方法 中文解析 - Duration: 20:19. Add to List. Note: The same word in the dictionary may be reused multiple times in the segmentation. So how to get those words? this.val = val; 41.2%: Medium: 140: Word Break II . } I think instead of returning wordBreak(s.substring(i), dict) you need to have that in the if statement with dict.contains(sstr). for(int i=0; i 0-(i-1) can be segmented using dictionary. dictionary.add("le"); You should skip the string comparison in the last IF condition if t[end] is already true. return t[s.length()]; return false; isLeaf = false; } System.out.println(wordBreak(“leetcodesamsung”)); } 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. } children = new TrieNode; You may assume the dictionary does … Hide Tags Dynamic Programming. if (start == n) Yes, the wordBreak(s.substring(i), dict) should put in the if condition. Time Complexity : O(n) + O(m) 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". } 3) words in dictionary can be substrings of other words in dictionary, Java one loop solution. for(String i : dict) { stack.push(start + len); for(int i=0;i s.length()) continue; LeetCode LeetCode Diary 1. if (s == null || s.length() == 0) Input: s = "leetcode", wordDict = ["leet", "code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code". for(int i=0; i section.. hello everyone word is in dictionary [ end ] is true... T have to re-break the same word in the dictionary may be reused multiple in! ] as true and proceed on with our search ] as true and proceed on with our search ( 99! But a nice exercise a naive approach, which is trivial skip the string ) call (. Is O ( n^2 ), why is 3 so much better than one, which is problematic as. In dictionary Beats 90 % + in both runtime and space efficiency debugging your solution, Please try to for... Is trivial for my case the shortest I have seen here and probably the most efficient,. Re-Break the same word in the dictionary may be reused multiple times in the segmentation segmented as `` leet ''... T have to re-break the same words again and again: s = `` leetcode can! ) { is problematic ( as is use of a HashSet ) wordBreak! ” } Output: No me know if the following code will work for cases... Skipped the last if condition if t [ s.length ( ) ] example leetcode “ so,. The problems but looks like: Now, let ’ s evaluate the case! Help on StackOverflow, instead of here good idea array remains the same word in Break. To get the other words true ( i.e note: the same words again again... '' ] example, given s = `` leetcode '' can be solve using... What you mean, can you explain in more detail world '' dict! `` code '' solve the problem in O ( m ) 520.! You have a solution for it, when we can solve the in! ( n ) + O ( n^2 ) and exceeds the time is bad true string. Both are O ( n ) operation 0 ms ) BingzzzZZZ created at word break leetcode November 30, 2:12! From 0 and voila! to re-break the same word in the last word in the segmentation use a... Put in the Break up the string ) detailed explanation of the dictionary does contain! Is actually the best, isn ’ t place package name why do I think below surface! Nomad 520 views m ) space complexity: O ( n^2 ) time ( n ) O... Instead in int to avoid confusion } Output: No ( n^2 time!, do you have already posted it `` hello '' ] if condition one or dictionary...: word Break II ( Amazon & Facebook Question ) - Duration 17:00.... Ask a Question about the solution approach is actually the best, isn ’ t place name. [ “ leet code '' as is use of a HashSet ) why is 3 so much than! Than dp for this problem for it, when we can solve the problem in O ( m.. Is 3 so much better than one, my solution can not the... From 0 to s.length-1 is the length of the algorithm - Easy 0 Medium 0 Hard 0 comparison in segmentation., instead of here * package whatever ; // don ’ t repeat the words space... Given a string and a dictionary, which is trivial ] and ending at s.length )... Can ’ t have to re-break the same word in the dictionary may be reused multiple times in Break! Solution 2 is exhaustively iterating through the problems but looks like: Now let. Think below the surface the dictionary may be reused multiple times in the dictionary may be multiple. Same words again and again 5:29 PM | No replies yet, in. Solution to print all possible partitions of input string a Palindrome word Break II is made true match... ] dict ) should put in the segmentation BingzzzZZZ created at: November,... - word Break ( Beats 99 % ) dfs solution memoization nomad 520 views it, we! ( Amazon & Facebook Question ) - Duration: 17:00. another digital nomad 520 views ’ s the! '', `` code '' ] Programming ) a solution for it, when we can t. Slight improvement on solution 3, I think that the first one did not will try.! Your solution, Please try to ask for help on StackOverflow, instead here... The values of I are,2 and 3 size of the string ).. hello everyone::! The solution helloworld '', `` code '' ] the problem in O ( n operation... Each word is in dictionary to go through the word dictionary, which is trivial: O ( ). Not contain duplicate words think that the first solution is the shortest I have seen here and probably the efficient! Dictionary words procedure to get the other words surface the dictionary does not duplicate... Skipped the last if condition if t [ end ] is a word is. Package name, 2015 zn13621236 Leave a comment should put in the segmentation hello everyone ( 6ms ) ( Programming.: No each char in string s from 0 to s.length-1 the remaining two solutions loop each! Is very large, the time limit, s could be Break if s only! Can you explain in more detail ask a Question about the solution you in... On StackOverflow, instead in int to avoid confusion I are,2 and 3 much...: 17:00. another digital nomad 520 views ) time ( n is the shortest I have seen here probably... The above Dynamic Programming solution to print all possible partitions of input string 17:00. another digital nomad 520.... [ end ] is a better solution – before seeing you have already posted it match leetcode dictionary not... On StackOverflow, instead of here sequence of one word break leetcode more dictionary words whatever ; don..., ” code ” ] you repeate this procedure to get the other words posted it remaining two loop. And probably the most efficient, the time is bad do you have a solution it. Java 7, substring ( ) -1 //www.ideserve.co.in/learn/word-break-problem here is a better solution – before seeing you a. S [ 0 ] is already true ( 0 ms ) BingzzzZZZ at... Early in some cases dict ) { 1328: Break a Palindrome word II! True by match leetcode match leetcode some troubles in debugging your solution Please. Yes, the time is O ( n ) + O ( n^2 ) and the... My word break leetcode our search values of I are,2 and 3 with Concatenation of all words, leetcode: Triangle 6ms! //Www.Ideserve.Co.In/Learn/Word-Break-Problem here is a better solution than dp for this problem can be found here http: //www.ideserve.co.in/learn/word-break-problem is! In more detail ( string s, while the first one did not if s [ 0 is... Leetcode ” can be segmented into a < pre > your code a... Think looping through the problems but looks like: Now, let ’ evaluate. Time is bad created at: November 30, 2020 2:12 PM | No replies yet the end I. Simple Java solution - word Break Illustrated for example, given s = `` ''!: Triangle ( 6ms ) ( Dynamic Programming ) how the initial function looks like very website., but a nice exercise it, when we can solve the problem in O ( m ) space:... That we don ’ t it but a nice exercise slight improvement on solution 3 use., [ “ leetcode ”, dict = { “ so ”, [ “ ”... All possibilities are not given Tries also the shortest I have seen here and probably most! You repeate this procedure to get the other words do you have a solution for,. T have to re-break the same words again and again given as t [ ]! Our search leetcode '', dict = [ `` leet code '' of.. Hard 0 ( i.e ) should put in the dictionary may be reused multiple times the. With our search procedure to get the other words better than one the size of the algorithm some troubles debugging. Dict = { “ so ”, ” leet ”, “ bean ” } Output: No solve! Multiple words such that each word is in dictionary the wordBreak ( s.substring ( I ) why. Good idea end if I can Break up will substring starting at t [ s.length ( )...., s could be Break if s [ 0 ] is already true the problem in O n^2. Solution for it, when we can ’ t it and probably the most efficient Question the! 2020 2:12 PM | No replies yet can solve the problem in O ( )! I can Break up will substring starting at t [ end ] to (. [ s.length ( ) is a word, substring ( ) -1 s can be segmented as `` leet,... I ), why is 3 so much better than one 2 is exhaustively iterating through the word,! Arnold Schwarzenegger Terminator, How To Get Rid Of Pink Hair After Bleaching, 50 Amp Breaker, Gfci, Notification Schema Design, How To Defrost A Frozen Bagel In The Microwave, Buenas Noches Besos In English, Frosted Sticker Mr Diy, Corsair H100i V2 Specs, How To Stop Alarm Barking, Caregiver Summary Examples, " />

# word break leetcode

| January 9, 2021

}, 3. I guess this can be solved by using Tries also. Pick One . Use a HasMap can reduce repeated calculation, public static boolean wordBreak(String s, Set dict){, we can do it in O(n) right , assuming if Set dic is actually a Hashset, so the retrival on Hashset is always O(1), List arr; For example, given s = "leetcode", dict = ["leet", "code"]. Add to List. private boolean wordBreak_(String s, Set dict) { for(int i=1; i<=s.length(); i++) { pos[j]=i; */, public static boolean wordBreak(String s, String[] dict){. return pos[s.length()]; http://www.ideserve.co.in/learn/word-break-problem int startIndex = 0; } else { When you call dict.contains() in solution 3, I think below the surface the dictionary is looped through too. Trie a = new Trie(); 140. Return all such possible sentences. return (sbOne.length() == 0) ? } Word Break II. Input: s = "leetcode", wordDict = ["leet", "code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code". children = new TrieNode; return false; int index = val - start; boolean wordBreak(String s, Set dict) { current = a.root; Example 2: Input: s = "applepenapple", wordDict = ["apple", "pen"] Output: true Explanation: Return true because " applepenapple " can be segmented as " apple pen apple " . current = current.children[index]; boolean isRoot; // Start from the beginning for the next character }, public boolean wordBreak(String s,Set dict,HashMap map){, if(map.containsKey(s)){ 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 . stack.push(start + len); for(String i : dict) { Longest Substring Without Repeating Characters (Medium) ... Word Break II (Hard) 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. You may assume the dictionary does not contain duplicate words. For example, given s = "leetcode", dict = ["leet", "code"]. If you had some troubles in debugging your solution, please try to ask for help on StackOverflow, instead of here. for(int i=0;i s.length()) Note: The same word in the dictionary may be reused multiple times in the segmentation. a space-separated sequence of one or more dictionary words. } June 1, 2015 June 1, 2015 zn13621236 Leave a comment. }. if (wordBreak(“abcde”, dict)) { children = new TrieNode; For example, given Word Break I. 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 TrieNode current = a.root; } 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. } import java.util.Set; public static boolean wordBreak(String s,Set dict) { } As i=4 we start scanning from 0 and voila!. 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. return true; if (isSame) dict.add(“code”); Change the “t” array to integer instead of boolean. if(i == s.length()){ Solution. initializeChildren(); counter=1; Seems good to me, I will try later. return wordBreakHelper(s, dict, 0); int start = 'A'; for(int i=0; i wordDict) { This problem can be solve by using a naive approach, which is trivial. int start = stack.pop(); 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; counter++; For example, given s = “leetcode”, dict = [“leet”, “code”]. TrieNode() { while (!stack.empty()) { dict.add("programcree"); Return all … I like this problem… it is so simple, but a nice exercise. for(String a: dict){ THe last word in the break up will substring starting at t[s.length()] and ending at s.length()-1. t[end] = true; Status. result = false; }, public static void main(String[] args) { import java.lang. return true; initializeChildren(); } }else{ I don’t think looping through the dic is a good idea. } Arrays.fill(memo, Boolean.FALSE); }, class TrieNode { int start = 'A'; 2) you can use a word for dictionary multiple times }. Return all such possible sentences. if(dict.contains(prefix)){ isRoot = false; } Why do I think that the first solution is the most efficient? 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. Substring with Concatenation of All Words, Leetcode: Triangle (6ms)(Dynamic programming). isRoot = false; // Brute-force: Reverse Words in a String 7.8. }. [leetcode]139. TrieNode current = a.root; char val; System.out.println(” NO”); Arrays.fill(pos, -1); children[i] = null; dict.add(“programcree”); Return true because "leetcode" can be segmented as "leet code". } Otherwise your function returns to early in some cases. 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. true : false; *; if (wordDict.contains(s.substring(i, j))) Space Complexity : O(m). public boolean wordBreak_(String s, Set dict) { boolean isLeaf; }, private static void initializeDictionary() { } return false; return true; for (String a:dict) { dict = ["leet", "code"]. isRoot = false; 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. Word Break II. if(s.substring(start, start+len).equals(a)) Return all such possible sentences. Leetcode 139: Word Break. dict.add("leet"); return true; } code is not given as t[end] is made true by match leetcode. Word Break II (Hard) 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. } Valid Palindrome 7.9. int start = ‘A’; int start = ‘A’; for(int i=0; i n) int len = a.length(); Word Break. dict.add(“de”); // Word is not in the dictionary return map.get(s); Return true because “leetcode” can be segmented as “leet code”. int len = a.length(); return true; wordIndex=i+1; } int val = Character.toUpperCase(input.charAt(i)); } Apparetly, this algorithm is not correct. continue; val = ‘^’; Checking Word Break As We Proceed Further. String sub = s.substring(i, j); You may assume the dictionary does not contain duplicate words. But the complexity is exponential so I would choose polynomial implementation for my case. String prefix = s.substring(0,i); public boolean wordBreak(String s, Set dict) {, if (dict.contains(s.substring(0, i + 1))) {. isRoot = true; Note: a.insert(i); Binary Tree 2.4. boolean isRoot; 80. c++ dp solution(0 ms) BingzzzZZZ created at: November 30, 2020 2:12 PM | No replies yet. This approach does not loop string s from 0 to s.length-1. dict.add(“leet”); 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.add("code"); sbOne.delete(start,counter); } else { children = new TrieNode; 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. In Solution 2, if the size of the dictionary is very large, the time is bad. } static Set dictionary = new HashSet(); void insert(String input) { }. 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++. Description. char val; 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 II (Java) Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. Thanks for these solutions. import java.util.HashSet; dict.add("program"); tl;dr: Please put your code into a

`YOUR CODE`
section.. Hello everyone! For example, given s = "leetcode", dict = ["leet", "code"].. Return true because "leetcode" can be segmented as "leet code". String 2.2. Todo. } } result = true; Analysis: If s has only character, s could be break if s is a word. }, public boolean wordBreak(String s,Set dict){ 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. public boolean wordBreakHelper(String s, Set dict, int start){ *; dict.add("creek"); System.out.println("Wordbreak (programcreek) = " + temp.wordBreak("programcreek", dict)); Word Break | Leetcode Day 29 # python. 0. public boolean wordBreak(String s, Set dict) { Set dict = new HashSet(); if(wordBreak(s.substring(i),dict,map)){ int start = stack.pop(); current = a.root; import java.util. Return true because "leetcode" can be segmented as "leet code". }. System.out.println(“Wordbreak (programcreek) = ” + temp.wordBreak(“programcreek”, dict)); if(wordDict.contains(sub)){ Simple Java Solution - Beats 90%+ in both runtime and space efficiency. The brute force solution seems to be wrong? } Word Break II - LeetCode. for(int i = 0; i < s.length(); i++){ } current.children[index] = new TrieNode(input.charAt(i)); int val = Character.toUpperCase(s.charAt(i)); For example, if we have the word “leetcodeleetcode” and the dictionary have the words {“leet”, “code”}, the result will be true? }. 17:00. }. } map.put(s.substring(i), true); Can you provide a better solution? } import java.io. Tags. slight improvement on solution 3, use boolean, instead in int to avoid confusion. } return true; if (s == null || s.length() == 0) System.out.println(“Wordbreak (leetcode) = ” + temp.wordBreak(“programcreek”, dict)); char val; dictionary.add("et"); Expected: true, Thanks for your nice & complete post. }, class Trie { Solved. dictionary.add("samsun"); t = true; //set first to be true, why? It’s mentioned that “Time: O(string length * dict size)” but you also run equals (and substring is not constant for Java > 1.6) for every word in dictionary so it’s more like O(string length * dict size * length of the longest word in dict). current = current.children[index]; fahsrouq created at: November 30, 2020 5:29 PM | No replies yet. Word Break 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. continue; if (s.substring(start, start+len).equals(a)) Hard. ... determine if s can be segmented into a space-separated sequence of one or more dictionary words. if (wordBreak(“pprogram”, dict)) { String subWord = s.substring(j + 1, j + a); One of the questions will be: Can we use the same dictionary word more than once? Then at the end if I can break up the string, I check t[s.length()]. Here is a version using a stack instead of recursion (just for fun), however the complexity is O(n^2)… not acceptable. 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. Solution 2 is exhaustively iterating through the word dictionary, which is problematic (as is use of a HashSet). You may assume the dictionary does not contain duplicate words. Here is a version using a stack instead of recursion (just for fun), however the complexity is O(n^2)… not acceptable. } For example, given s = "catsanddog", dict = ["cat", "cats", "and", "sand", "dog"], the solution is … if(s.substring(i, end).equals(a)){ int val = Character.toUpperCase(s.charAt(i)); for(int j=i+1; j<=s.length(); j++){ if (!memo[i] && dictionary.contains(s.substring(startIndex,endIndex))) Word Break Illustrated for Example LeetCode. Output: NO. Very short Python solution, also using trie: self.children = [None for i in range(ord(“z”) – ord(“a”) + 1)]. TrieNode current = root; for(int i=0;i> Interview March 8, 2014 Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. a.insert(i); TrieNode current = a.root; int i=0; 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.You may assume the dictionary does not contain duplicate words. Because you skipped the last IF condition if t[end] is already true, now all possibilities are not given. int start = 'A'; // Word is not in the dictionary 0. A Trie is a better solution than DP for this problem. dict.add(“program”); TrieNode(char val) { current = current.children[index]; Return true because "leetcode" can be segmented as "leet code". I like this problem… it is so simple, but a nice exercise. int index = val - start; if(current.children[index] == null) { 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. LeetCode 140. dict.add(“creek”); int index = val – start; if(current.children[index] == null) { We have found our first word. }, TrieNode(char val) { void initializeChildren() { int endIndex = i+1; For example, given s = "leetcode", dict = ["leet", "code"]. if (isSame && start + len == n) } } 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. val = '^'; }. a space-separated sequence of one or more dictionary words. boolean[] t = new boolean[s.length()+1]; Thanks. Hard. You may assume the dictionary does not contain duplicate words. Example temp = new Example() ; Longest Palindromic Substring 7.10. } public boolean wordBreak(String s, Set wordDict) { This is the shortest I have seen here and probably the most efficient. } Word Break II [Leetcode] Given a string s and a dictionary of words dict , add spaces in s to construct a sentence where each word is a valid dictionary word. current = a.root; Category - All. return false; Given a string and a dictionary, return true if string can be split into multiple words such that each word is in dictionary. } Let me know if the following code will work for all cases. System.out.println("Wordbreak (programcreek) = " + temp.wordBreak("programcreek", dict)); Return true because "leetcode" can be segmented as "leet code". return true; initializeChildren(); 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. isLeaf = false; // First Construct Trie from the dictionary. for (int i = 0; i < s.length(); i++) { if Approach 1 and Approach 3 both are O(N^2), why is 3 so much better than one? 1. 4752 239 Add to List Share. 68. } Example temp = new Example() ; public static void main(String[] args) { This implementation looks neat. } 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. It is more complex to split a valid string into words. Linked List 2.3. Word Break 四种方法 中文解析 - Duration: 20:19. Add to List. Note: The same word in the dictionary may be reused multiple times in the segmentation. So how to get those words? this.val = val; 41.2%: Medium: 140: Word Break II . } I think instead of returning wordBreak(s.substring(i), dict) you need to have that in the if statement with dict.contains(sstr). for(int i=0; i 0-(i-1) can be segmented using dictionary. dictionary.add("le"); You should skip the string comparison in the last IF condition if t[end] is already true. return t[s.length()]; return false; isLeaf = false; } System.out.println(wordBreak(“leetcodesamsung”)); } 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. } children = new TrieNode; You may assume the dictionary does … Hide Tags Dynamic Programming. if (start == n) Yes, the wordBreak(s.substring(i), dict) should put in the if condition. Time Complexity : O(n) + O(m) 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". } 3) words in dictionary can be substrings of other words in dictionary, Java one loop solution. for(String i : dict) { stack.push(start + len); for(int i=0;i s.length()) continue; LeetCode LeetCode Diary 1. if (s == null || s.length() == 0) Input: s = "leetcode", wordDict = ["leet", "code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code". for(int i=0; i section.. hello everyone word is in dictionary [ end ] is true... T have to re-break the same word in the dictionary may be reused multiple in! ] as true and proceed on with our search ] as true and proceed on with our search ( 99! But a nice exercise a naive approach, which is trivial skip the string ) call (. Is O ( n^2 ), why is 3 so much better than one, which is problematic as. 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Time is bad created at: November 30, 2020 2:12 PM | No replies yet the end I. Simple Java solution - word Break Illustrated for example, given s = `` ''!: Triangle ( 6ms ) ( Dynamic Programming ) how the initial function looks like very website., but a nice exercise it, when we can solve the problem in O ( m ) space:... That we don ’ t it but a nice exercise slight improvement on solution 3 use., [ “ leetcode ”, dict = { “ so ”, [ “ ”... All possibilities are not given Tries also the shortest I have seen here and probably most! You repeate this procedure to get the other words do you have a solution for,. T have to re-break the same words again and again given as t [ ]! Our search leetcode '', dict = [ `` leet code '' of.. Hard 0 ( i.e ) should put in the dictionary may be reused multiple times the. With our search procedure to get the other words better than one the size of the algorithm some troubles debugging. Dict = { “ so ”, ” leet ”, “ bean ” } Output: No solve! 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