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  1. www.geeksforgeeks.org › problems › word-ladderWord Ladder I | Practice | GeeksforGeeks

    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

  2. www.geeksforgeeks.org › problems › word-ladder-iiWord Ladder II | Practice | GeeksforGeeks

    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.

  3. practice.geeksforgeeks.org › problems › word-ladderPractice | GeeksforGeeks | A computer science portal for geeks

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  4. www.geeksforgeeks.org › word-ladder-length-ofWord Ladder (Length of shortest chain to reach a target word)

    Mar 2, 2023 · 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.

  5. leetcode.com › problems › word-ladderWord Ladder - LeetCode

    Example 1: Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"] Output: 5 Explanation: One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.

  6. practice.geeksforgeeks.org › problems › word-ladderGeeksforGeeks

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  7. leetcode.com › problems › word-ladder-iiWord Ladder II - LeetCode

    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.