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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.
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.
Given two distinct words startWord and targetWord, and a list denoting wordList of unique words of equal lengths. Find all shortest transformation sequence(s) from startWord to targetWord. You can return them in any order possible.Keep&nbs.
4 days ago · Given a dictionary, and two words ‘start’ and ‘target’ (both of same length). Find length of the smallest chain from ‘start’ to ‘target’ if it exists, such that adjacent words in the chain only differ by one character and each word in the chain is a valid word i.e., it exists in the dictionary.
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.
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.
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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:
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 words in wordList are unique. Solution (javascript) /** * @param {string} beginWord . * @param {string} endWord . * @param {string[]} wordList . * @return {number} . */ var ladderLength = function(beginWord, endWord, wordList) { var wordSet = new Set (wordList); var queue = []; var step = 0 ; var word = '' ;
