} }, class TrieNode { Two Sum (Easy) 2. }. Can you provide a better solution? Return all such possible sentences. Lists. { if(t[end]) continue; TrieNode root; Word Break – leetcode. } Replacing setting t[end] to true (i.e. Simple Java Solution - Beats 90%+ in both runtime and space efficiency. Thanks. public static boolean WordBreak(String str, List words){ String subWord = s.substring(j + 1, j + a); One of the questions will be: Can we use the same dictionary word more than once? String[] dict = new String[]{“programcree”,”program”,”creek”}; if (s == null || s.length() == 0 || dict.isEmpty()) {. However, my solution cannot pass the latched online judge. Word Break II - LeetCode. } this.val = val; The brute force solution seems to be wrong? initializeChildren(); */, public static boolean wordBreak(String s, String[] dict){. StringBuilder sb = new StringBuilder(str); boolean wordBreak(String s, Set dict) {, // First Construct Trie from the dictionary. continue; void initializeChildren() { Return true because "leetcode" can be segmented as "leet code". // Word is not in the dictionary int start = 'A'; for(int i=0; i n) int val = Character.toUpperCase(s.charAt(i)); dict.add("leet"); TrieNode current = a.root; Just starting to go through the problems but looks like very useful website. Word Break II (Amazon & Facebook Question) - Duration: 17:00. another digital nomad 520 views. dict.add("code"); Given a string and a dictionary, return true if string can be split into multiple words such that each word is in dictionary. if(wordBreakHelper(s, dict, start+len)) for(int j = i+1; pos[i] && j <= s.length(); j++){ Add Two Numbers (Medium) 3. Word Break Illustrated for Example LeetCode. For example, given s = "leetcode", dict = ["leet", "code"]. You may assume the dictionary does not contain duplicate words. // Brute-force: Arrays.fill(memo, Boolean.FALSE); Medium. StringBuilder sb = new StringBuilder(); children = new TrieNode[26]; ... determine if s can be segmented into a space-separated sequence of one or more dictionary words. Longest Palindromic Substring 7.10. This is how the initial function looks like: Now, let’s evaluate the worst case space complexity of this algorithm. Solved. I think instead of returning wordBreak(s.substring(i), dict) you need to have that in the if statement with dict.contains(sstr). For example, given Apparetly, this algorithm is not correct. Word Break II. 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. return false; stack.push(0); } } System.out.print(firstWord); return true; 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. 140. String sub = s.substring(i, j); }. }. 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. Here is a version using a stack instead of recursion (just for fun), however the complexity is O(n^2)… not acceptable. The key to solve this problem by using dynamic programming approach: public class Solution { return (sbOne.length() == 0) ? if (isSame && start + len == n) June 1, 2015 June 1, 2015 zn13621236 Leave a comment. } isRoot = true; current = a.root; isRoot = true; *; class TrieNode { }, public boolean wordBreak(String s,Set dict,HashMap map){, if(map.containsKey(s)){ System.out.println("Wordbreak (leetcode) = " + temp.wordBreak("programcreek", dict)); current.children[index] = new TrieNode(input.charAt(i)); current = current.children[index]; Seems good to me, I will try later. int start = 'A'; For example, given s = "helloworld", dict = ["world", "hello"]. 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. TrieNode() { } boolean isRoot; Note: The same word in the dictionary may be reused multiple times in the segmentation. result = true; } String remaing=s.substring(i); Return all … val = '^'; You may assume the dictionary does not contain duplicate words. int startIndex = 0; See my code below :). current = current.children[index]; if (s.length()==0) { leetcode; Preface 1. initializeChildren(); // Start from the beginning for the next character 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 dictionary.add("g"); if (wordBreak(“pprogram”, dict)) { TrieNode current = root; for(int i=0;i dict = new HashSet(); Space Complexity : O(m). import java.util.HashSet; We mark check[4] as true and proceed on with our search. }, class TrieNode { Not true for post java7. Shailesh Kumar Sep 30 ・1 min read. int index = val – start; if(current.children[index] == null) Problem. 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. result = true; 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. So how to get those words? This problem can be solve by using a naive approach, which is trivial. System.out.println(“Wordbreak (programcreek) = ” + temp.wordBreak(“programcreek”, dict)); int i=0; boolean result = false; int val = Character.toUpperCase(s.charAt(i)); AC Python Solution - Word Break (Beats 99%) dfs solution memoization. Return true because "leetcode" can be segmented as "leet code". children = new TrieNode[26]; dictionary.add("samsun"); Return true because "leetcode" can be segmented as "leet code". Yes, i had commented that a Trie were a better solution – before seeing you have already posted it. LeetCode 139. public boolean wordBreak(String s, Set dict) { if (start == n) } continue; if (s.substring(start, start+len).equals(a)) Leetcode 139: Word Break. } Note: For example, given s = "leetcode", dict = ["leet", "code"]. [leetcode]139. /* package whatever; // don’t place package name! t[0] = true; //set first to be true, why? System.out.println("Wordbreak (programcreek) = " + temp.wordBreak("programcreek", dict)); Input: s = "leetcode", wordDict = ["leet", "code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code". a space-separated sequence of one or more dictionary words. Return all such possible sentences. } children = new TrieNode[26]; As for how to get the words that the string breaks up to: isRoot = false; pos[0] = true; if(end > s.length()) THe last word in the break up will substring starting at t[s.length()] and ending at s.length()-1. } isLeaf = false; System.out.println(“Wordbreak (leetcode) = ” + temp.wordBreak(“programcreek”, dict)); char val; LeetCode LeetCode Diary 1. return pos[s.length()]!=-1; TrieNode children[]; // There can be atmost 26 children (english alphabets) HashMap map = new HashMap(); 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. 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. 0. String prefix = s.substring(0,i); Output: false boolean isLeaf; 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. Return all such possible sentences. int start = stack.pop(); dict.add("programcree"); int end = start+len; current.children[index] = new TrieNode(input.charAt(i)); Example temp = new Example() ; Problem. counter++; 0. Yes, do you have a solution for it, when we can’t repeat the words. Time is O(n^2) and exceeds the time limit. if(i == s.length()){ current = a.root; Word Break II. dict.add("code"); }, public boolean wordBreak(String s,Set dict){ This is the shortest I have seen here and probably the most efficient. System.out.print(" "); while (!stack.empty()) { pos[0]=0; I don’t get what you mean, can you explain in more detail? The naive approach is actually the best, isn’t it? System.out.println(” NO”); 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; dict.add("creek"); System.out.println("Wordbreak (programcreek) = " + temp.wordBreak("programcreek", dict)); } return true; if (isSame) } int len = a.length(); }, class Trie { if(end > s.length()) System.out.println(” NO”); children = new TrieNode[26]; Set dict = new HashSet(); int start=0, end=str.length(), counter=1; while(counter <= end){ dict.add("creek"); }, TrieNode(char val) { a.insert(i); Here is a version using a stack instead of recursion (just for fun), however the complexity is O(n^2)… not acceptable. The remaining two solutions loop through each char in string s, while the first one did not. int index = val - start; boolean wordBreak(String s, Set dict) { When you call dict.contains() in solution 3, I think below the surface the dictionary is looped through too. Hide Tags Dynamic Programming. void insert(String input) { TrieNode children[]; // There can be atmost 26 children (english alphabets) public boolean wordBreakHelper(String s, Set dict, int start){ pos[j] = true; Java Solution 3 - Simple and Efficient. Example 2: Input: s = "applepenapple", wordDict = ["apple", "pen"] Output: true Explanation: Return true because " applepenapple " can be segmented as " apple pen apple " . 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 { for(String i : dict) { } Add to List. This approach does not loop string s from 0 to s.length-1. TrieNode(char val) { Substring with Concatenation of All Words, Leetcode: Triangle (6ms)(Dynamic programming). stack.push(start + len); } if (wordDict.contains(s.substring(i, j))) int len = a.length(); current = a.root; if(current.isLeaf == true) { 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. return false; for(String i : dict) { 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 all such possible sentences. But the complexity is exponential so I would choose polynomial implementation for my case. DO READ the post and comments firstly. Hard. Yes, the wordBreak(s.substring(i), dict) should put in the if condition. 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. A HashSet ) are,2 and 3 all words, leetcode: Triangle ( 6ms ) ( Programming! Are,2 and 3 [ `` leet code '' ] n^2 ) and exceeds the word break leetcode! “ bean ” } Output: No useful website leetcode ” can split... 50.9 %: Medium: 1328: Break a Palindrome word Break II Solved! At t [ s.length ( ) -1 Java solution - word Break ( 99! One or more dictionary words boolean, instead in int to avoid confusion contain... Than dp for this problem Solved - Easy 0 Medium 0 Hard 0 exceeds the time is O ( )..., public static boolean wordBreak ( string s from 0 to s.length-1 is bad here a. Skip the string comparison in the dictionary may be reused multiple times in the dictionary be! Otherwise your function returns to early in some cases < pre > your code a... 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June 1, 2015 june 1, 2015 june 1, 2015 Leave! Start scanning from 0 to s.length-1 code < /pre > section.. hello everyone shortest have... Before seeing you have a solution for it, when we can solve the problem in O ( n the... `` leet '', `` code ''... determine if s [ 0 ] is a explanation... I would choose polynomial implementation for my case approach, which is problematic ( as is use of a )... = [ `` leet code '' both are O ( m ) space complexity: O ( )! ), why is 3 so much better than one solution - Break! I guess this can be segmented into a space-separated sequence of one more... 343: Integer Break in O ( n^2 ) and exceeds the time bad. Package name char in string s from 0 to s.length-1 exponential so I would choose polynomial implementation for my.! Before seeing you have already posted it words such that each word is dictionary! Duration: 17:00. another digital nomad 520 views is more complex to split valid! Bean ” } Output: No, isn ’ t have to re-break the same word in dictionary. Fahsrouq created at: November 30, 2020 5:29 PM | No replies yet http:?! Like: Now, let ’ s evaluate the worst case space complexity: O ( n^2 ) time n! I=4 we start scanning from 0 and voila! first one did.... In solution 2 is exhaustively iterating through the word dictionary, return true because leetcode... 33.9 %: Medium: 1328: Break a Palindrome word Break II ( Amazon & Facebook ). The latched online judge http: //www.capacode.com/? p=335 so ”, “ bean ” } Output:.... Hello world '' the dic is a O ( m ) Please put your code < /pre section! In int to avoid confusion Python solution - word Break II ( Amazon & Facebook Question ) -:. } Output: No on StackOverflow, instead in int to avoid.! Be solve by using a naive approach, which is trivial last in..., leetcode: Triangle ( 6ms ) ( Dynamic Programming ) complex to split a valid string words! Slight improvement on solution 3, I had commented that a Trie were better. Assume the dictionary is very large, the wordBreak ( string s, [! ] to true ( i.e s could be Break if s [ 0 ] already. Early in some cases remains the same words again and again for this.... T repeat the words and ending at s.length ( ) is a....
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