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Constraints: * 1 <= beginWord.length <= 10 * endWord.length == beginWord.length * 1 <= wordList.length <= 5000 * wordList [i].length == beginWord.length * beginWord, endWord, and wordList [i] consist of lowercase English letters. * beginWord != endWord * All the words in wordList are unique.
A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s 1-> s 2-> ... -> s k such that:. Every adjacent pair of words differs by a single letter. Every s i for 1 <= i <= k is in wordList.Note that beginWord does not need to be in wordList.; s k == endWord; Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 ...
Word Ladder II. Hard. A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s 1 -> s 2 -> ... -> s k such that: Every adjacent pair of words differs by a single letter. Every s i for 1 <= i <= k is in wordList.
class Solution {public int ladderLength (String beginWord, String endWord, List < String > wordList) {Set < String > wordSet = new HashSet <> (wordList); if (! wordSet. contains (endWord)) return 0; int ans = 0; Queue < String > q = new ArrayDeque <> (Arrays. asList (beginWord)); while (! q. isEmpty ()) {++ ans; for (int sz = q. size (); sz > 0 ...
Given two distinct words startWord and targetWord, and a list denoting wordList of unique words of equal lengths. Find the length of the shortest transformation sequence from startWord to targetWord.Keep the following conditions in m.
In-depth solution and explanation for LeetCode 127. Word Ladder in Python, Java, C++ and more. Intuitions, example walk through, and complexity analysis. Better than official and forum solutions.
Apr 5, 2016 · Solution 1: BFS. BFS minimum step model. This problem can be solved with naive BFS, or it can be optimized with bidirectional BFS to reduce the search space and improve efficiency. Bidirectional BFS is a common optimization method for BFS, with the main implementation ideas as follows:
Word Ladder. Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that: Only one letter can be changed at a time. Each transformed word must exist in the word list. Note that beginWord is not a transformed word. Note:
The Word Ladder LeetCode Solution – “Word Ladder” states that you are given a string beginWord, string endWord, and a wordList. We need to find the shortest transformation sequence length (if no path exists, print 0) from beginWord to endWord following the given conditions: All the Intermediate Words should be present in wordList.
Word Ladder. Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that: Only one letter can be changed at a time. Each transformed word must exist in the word list. Note that beginWord is not a transformed word. For example,
