Word Ladder Leetcode optimised 2 solutions c++ code and explanation

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord
Every adjacent pair of words differs by a single letter.
Every si is in wordList. Note that beginWord does not need to be in wordList.
sk == 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 if no such sequence exists.

[@nehasht2]

First thing when I come across questions ..I go through your playlist if not found then search elsewhere.....I really appreciate the way you explain

[@rahulkarnalk4479]

exactly doing the same

[@K-IT-VishalPanchal]

your explanation is amazing mostly your way to explain code with the help of example
✌✌

[@shalsteven]

THANK YOU so much for your clear and DETAIL explanation!

[@amangaur4231]

Too good 🔥 explaination....clearly understood 👍👍

[@sachinsacse]

Nice explaination...
BeginWord could be added to vector at starting itself before creating Adj list

[@NaveenKumar-os8dv]

thank you, as always your explanation is great!

[@code_lover2889]

Keep it up ma'am ❤️❤️

[@mahboobbiswas1304]

Your explanation to good.

[@vatsaldoshi7282]

Nice explanation! Thanks

[@Idukhan-jj9kc]

Nice explanation alli

[@aryannagraj4681]

you explain very well

[@rachitbadoni5717]

Nice explanation Di❤

[@dhananjoydey1337]

thank you so much

[@AmanSharma-vb5jl]

Thanks

[@MilindGupta]

Hey alisha cover some dp questions too if possible

[@ash11music]

can someone explain why time complexity is O(m*m*n) and not O(m*n) ?? it seems there are only 2 nested loops right?

[@GojoSaturo-qe7uo]

(first loop -> n)*( second loop -> m)*( diff function is called (which traverse the word)-> size of word)
n*m* size of word

[@shyren_more]

great explanation, could you please share the code?

[@HarendraKumar-yr4gt]

😍😍

[@aryanpatidar5392]

if possible, can you please share your code

[@veenurohan3267]

class Solution {
public:
int ladderLength(string beginWord, string endWord, vector& wordList) {
if (count(wordList.begin(), wordList.end(), endWord) == 0) return 0;
unordered_set Set;
queue Queue;
unordered_map Map;
wordList.push_back(beginWord);
int N = wordList.size(), M = beginWord.size();
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
string x = wordList[i].substr(0, j) + '*' + wordList[i].substr(j + 1, M - j - 1);
Map[x].push_back(wordList[i]);
}
}
Queue.push({beginWord, 1});
while (!Queue.empty()) {
pair P = Queue.front();
Queue.pop();
for (int j = 0; j < M; j++) {
string x = P.first.substr(0, j) + '*' + P.first.substr(j + 1, M - j - 1);
for (string &word: Map[x]) {
if (Set.find(word) != Set.end()) continue;
if (word == endWord) return P.second + 1;
Queue.push({word, P.second + 1});
Set.insert(word);
}
}
}
return 0;
}
};

[@mahboobbiswas1304]

hello, can you make videos for DSA?
how can contact you?

[@probabilitycodingisfunis1]

[e-mail]