Let's continue the habit of commenting “understood” if you got the entire video. Please give it a like too, you don't 😞 Do follow me on Instagram: striver_79
Honestly I feel that the hard part about this question was figuring out that it is a graph question. But since I was following the playlist I already knew that it is a graph question. Which is why upon looking the question the intuition came easily to me.
@@abhinavgupta7048 it is generally advised to not use unordered sets and maps, and due to collisions, the worst case time complexity to access elements becomes O(n)
thank you striver i was able to solve this problem on my own because of your graph playlist. heartiful thanks for this playlist.this is the best out there.
For second time I was able to solve GFG hard problem on my own , such a nice teaching. Though I doubt if such approach would've hit, if it hadn't been a part of Graph series.
One more approach can be making the graph first & then finding the shortest path in it. We can make a graph where there will be a edge between all those nodes who differ by only 1 character. The graph creation will take time = (N^2)*M (where N is length of words array & M is length of longest word). Once graph is made we can apply BFS & also handle the case where target cannot reach.
i tried this but it says tle?? class Solution(object): def ladderLength(self, beginWord, endWord, wordList): """ :type beginWord: str :type endWord: str :type wordList: List[str] :rtype: int """ def issimilar(s1,s2): if len(s1)!=len(s2): return False l=len(s1) flag=True for i in range(l): if s1[i]!=s2[i] and flag==False: return False if s1[i]!=s2[i]: flag=False return True if endWord in wordList: final=wordList.index(endWord) que=[]
n=len(wordList) adj=[[] for _ in range(n)] visited=[0 for _ in range(n)] for i in range(n-1): for j in range(1,n): if issimilar(wordList[i],wordList[j]): adj[i].append(j) adj[j].append(i) for i in range(n): if issimilar(wordList[i],beginWord): que.append((i,2)) visited[i]=1
while que: v,dis=que.pop(0) if v==final: return dis for i in adj[v]: if visited[i]==0: que.append((i,dis+1)) visited[i]=1 return 0
@@jyothiyadav2595 also created the graph first and it worked for me class Solution { class Pair { int n, d; Pair(int n ,int d) { this.n = n; this.d = d; } } boolean check(String st1, String st2) { int c=0; for(int i=0; i
I solved this question in the following way which gives me the time complexity of O(n*m). I basically created the graph using adj list method and performed normal BFS instead of checking words in the set after modifying each letter of word. class Solution { public: bool checkDiff(string a, string b){ int ptr=0, count=0; while(ptr!=a.size()){ if(a[ptr]==b[ptr]){ ptr++; continue; } ptr++; count++; } return count==1; } int ladderLength(string beginWord, string endWord, vector& wordList) { wordList.insert(wordList.begin(), beginWord); int n = wordList.size(); vector adj[n]; //Create undirected graph for(int i=0; i
Thank you so much Striver for explaining problems in a very understandable way. Kudos to you for achieving lot at a very young age in the world of Top-tech
I remember this problem came in Leetcode daily in Feb 2022. Just copied the code and did not understand anything. One year later I understood the code. Thank you for the explanation.
It took me around one day to solve these question using my method but in the end it got TLE on leetcode and now after seeing the solution the questions seems easy.
Hey @striver this solution is exceeding time limit in leetcode, so instead of assigning all letter from a-z , I tried with letters only in wordList and it worked. I just added below function to get all the letters into vector- private: vector letters(vector&wordList){ unordered_set st; vector result; for(auto it: wordList){ for(int i=0;i
My approach is different: This can be visualised as an undirected weighted graph with unit weights. We can created graph with word as each node and all neighbours are the words with only one character changed. Then apply exact same BFS as we implement in the previous video. No code changes at all. Quite simple and passes all testcases. bool isAdjacent(const string& word1, const string& word2) { int diff = 0; for (int i = 0; i < word1.size(); ++i) { if (word1[i] != word2[i]) { diff++; if (diff > 1) return false; } } return diff == 1; } // Function to create the adjacency list graph unordered_map buildGraph(vector& wordList) { unordered_map graph; // Add an empty list for each word for (const string& word : wordList) { graph[word] = {}; } // Connect each word with its adjacent words for (int i = 0; i < wordList.size(); ++i) { for (int j = i + 1; j < wordList.size(); ++j) { if (isAdjacent(wordList[i], wordList[j])) { graph[wordList[i]].push_back(wordList[j]); graph[wordList[j]].push_back(wordList[i]); } } } return graph; } int wordLadderLength(string startWord, string targetWord, vector& wordList) { // If targetWord is not in the wordList, no valid transformation is possible if (find(wordList.begin(), wordList.end(), targetWord) == wordList.end()) { return 0; } // Include the startWord in the word list if it's not already there if (find(wordList.begin(), wordList.end(), startWord) == wordList.end()) { wordList.push_back(startWord); } // Build the graph from the word list unordered_map graph = buildGraph(wordList); // Perform BFS to find the shortest path unordered_map distance; for (const string& word : wordList) { distance[word] = INT_MAX; // Initialize distances with infinity } distance[startWord] = 1; queue q; q.push(startWord); while (!q.empty()) { string currentWord = q.front(); q.pop(); // Traverse all neighbors of the current word for (const string& neighbor : graph[currentWord]) { if (distance[neighbor] == INT_MAX) { // If not visited distance[neighbor] = distance[currentWord] + 1; q.push(neighbor); // If we reach the target word, return the distance if (neighbor == targetWord) { return distance[neighbor]; } } } } // If BFS completes without finding the targetWord, return 0 return 0; }
Thanks striver, Was able to solve it by myself although my approach was different from this, I considered every node has a edge to every other node and each edge has a weight equal to the difference b/w the 2 vertices(words) Then just used the shortest distance for an undirected graph using BFS algo
Since its bfs, steps are going to be common for each level traversal. So no need to save the steps inside queue. Can maintain it outside & keep incrementing as we finish one traversal.
we can create graph from the word list and do topological sort on it. that will give only limited characters that exists in word list and we can try out replacing characters present in word list only. with topological sort the we get array as h,d,l,c,o,t,g. Now I know before o I have 4 characters to try for first place i.e. h,d,l,c. then for g I have 2 characters o and t with same BFS algoritm we can avoid trying out all a-z combination and also it makes sense logically because after transformation if word should be from word list then try out with characters that are only in wordlist. if z is not present in wordlist in any word why to try out with that. Considering if there is no cycle in it. This needs to be asked to interviewer whether there will be cycle in the graph. If cycle exists then the other approach is to iterate through all the wordlist and put them in set. Atleast you will be comparing less characters then a-z
Simple bruteForce can be : for each element(say "hit") in list, compare it with the all other elements. If the only 1 char varies that we can add as part of adjacency List of "hit" and so on.
Briliiantly explained When I read the question for the first time, I was so scared and I had no clue how to proceed. I was thing of sets and recursion but I did not even try coz Iwas scared. But u made it soo simple that after watching half of the video, I did it in 1 go . U r amazing striver. please continue to bring such playlists , its so helpful
One can figure out that it was a graph question by looking at src & dest strings, & shortest sequence of transformations were asked...So, some 'feeling' of shortest path from src (beginWord) to dest (endWord) comes...
best method to solve this question 😁thanks striver I got your algo and tried myself , just make a little mistake but it was corrected bcoz of your code
I have a different solution, i created a graph representation using adjacency list (word is the key, words it can transform into are list values), then applied shortest distance method using dijkstra's algorithm.
Instead of converting str to all the possible strings we can simply compare str with all the strings in word_List and see if the difference btw them is 1 or less and if so then add in queue to do BFS.
Striver, thank you so much for graph series, per please can you also make series on segement trees, tried learning from so many places, but nowhere satisfactory content is found, bana do na ek series please
once i knew it was a graph question, i tried this myself : class Solution { private: bool check(const string& w1, const string& w2) { int differences = 0; for (int i = 0; i < w1.length(); ++i) { if (w1[i] != w2[i]) { differences++; if (differences > 1) return false; } } return true; } public: int ladderLength(string beginWord, string endWord, vector& wordList) { bool possible = false; int index = -1; for (int i = 0; i < wordList.size(); i++) { if (wordList[i] == endWord) { index = i; possible = true; break; } } if (!possible) return 0;
wordList.insert(wordList.begin(), beginWord); int n = wordList.size(); vector adj(n); for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { if (check(wordList[i], wordList[j])) { adj[i].push_back(j); adj[j].push_back(i); } } } vector distance(n, INT_MAX); queue q; q.push({0, 0}); distance[0] = 0; while (!q.empty()) { int node = q.front().first; int dist = q.front().second; q.pop(); for (int val : adj[node]) { if (distance[val] > dist + 1) { distance[val] = dist + 1; q.push({val, dist + 1}); } } } if (distance[index + 1] < INT_MAX) return distance[index + 1] + 1; return 0; } };
A Small Optimization Can be MADE further ... So currently, for each index of the string we are traversing 26 times representing 26 characters / alphabets in the English Dictionary. But what if we exactly store the possible set of characters that are only available in wordList. There comes the optimization where we store set of all alphabets for each index & always iterate those in the while loop of queue.
really don't understand why this is marked as "hard", this is one of the easiest problems, and it can also be considered as tree too.. regardless.. thank you Striver❤
Hey Striver! Understood the concept thoroughly. Had one doubt, my first intuition was to use DP in this question instead of graph. SO why graph is more optimal then dp?
1) The question has a source to destination pattern, it's guiding you a path that you start from this word and end somewhere... This intends graph... Also they are saying "shortest" while giving you source and destination so you should think of shortest path algorithm 2) Reason why you thought of Recursion/DP is because we are trying every letter combination which is fair to think but you apply DP when you have a vast search space and you have no clue which can be the answer but here they've given you the dictionary list so they are telling you the search space
One different approach is make a graph with edges between hit and hot, hot and dot, hot and dot, hot and lot, .... And now you have to find the shortest distance between two points (beginWord and endWord) using BFS.
Let's continue the habit of commenting “understood” if you got the entire video. Please give it a like too, you don't 😞
Do follow me on Instagram: striver_79
Honestly I feel that the hard part about this question was figuring out that it is a graph question. But since I was following the playlist I already knew that it is a graph question. Which is why upon looking the question the intuition came easily to me.
naah since,src and dest pointers were there, it was easy to figure out it was a graph,but rest was difficult
How is this even a Graph problem?
so us bro🫂🫂same thought
same thought
Watching your lectures before the Google on-site interview
if you got selected refer me too :)
linkedln id dedo bhaiya XD
I have my google on-site next month.. want to discuss on prep so can we connect on LinkedIn ?
how did it go?
Ok Googler
The patience with which you do dry run is really commendable!!!!🔥
Instead of erasing the string from the set, you can take unordered_map and mark it as true which is generally used in BFS for marking visited nodes.
ya but that will unnecessarily increase the space complexity
Terrible idea as it not only increases SC , but also increases COLLISIONS leading to worst case in hashmap
@@jagannathan1014 can you please elaborate why it is a terrible idea and how will it increase collisions
yeah you can do it too...
@@abhinavgupta7048 it is generally advised to not use unordered sets and maps, and due to collisions, the worst case time complexity to access elements becomes O(n)
Understood. Graph is slightly tough, but you are making it easy for us thank you🤘.
thank you striver i was able to solve this problem on my own because of your graph playlist. heartiful thanks for this playlist.this is the best out there.
For second time I was able to solve GFG hard problem on my own , such a nice teaching. Though I doubt if such approach would've hit, if it hadn't been a part of Graph series.
One more approach can be making the graph first & then finding the shortest path in it. We can make a graph where there will be a edge between all those nodes who differ by only 1 character. The graph creation will take time = (N^2)*M (where N is length of words array & M is length of longest word). Once graph is made we can apply BFS & also handle the case where target cannot reach.
I was thinking the same way
i tried this but it says tle??
class Solution(object):
def ladderLength(self, beginWord, endWord, wordList):
"""
:type beginWord: str
:type endWord: str
:type wordList: List[str]
:rtype: int
"""
def issimilar(s1,s2):
if len(s1)!=len(s2):
return False
l=len(s1)
flag=True
for i in range(l):
if s1[i]!=s2[i] and flag==False:
return False
if s1[i]!=s2[i]:
flag=False
return True
if endWord in wordList:
final=wordList.index(endWord)
que=[]
n=len(wordList)
adj=[[] for _ in range(n)]
visited=[0 for _ in range(n)]
for i in range(n-1):
for j in range(1,n):
if issimilar(wordList[i],wordList[j]):
adj[i].append(j)
adj[j].append(i)
for i in range(n):
if issimilar(wordList[i],beginWord):
que.append((i,2))
visited[i]=1
while que:
v,dis=que.pop(0)
if v==final:
return dis
for i in adj[v]:
if visited[i]==0:
que.append((i,dis+1))
visited[i]=1
return 0
can someone help??
@@jyothiyadav2595
also created the graph first and it worked for me
class Solution {
class Pair {
int n, d;
Pair(int n ,int d) {
this.n = n;
this.d = d;
}
}
boolean check(String st1, String st2) {
int c=0;
for(int i=0; i
i just watched the logic you have used and i was able to write code by myself , Thanks..🙏
I solved this question in the following way which gives me the time complexity of O(n*m). I basically created the graph using adj list method and performed normal BFS instead of checking words in the set after modifying each letter of word.
class Solution {
public:
bool checkDiff(string a, string b){
int ptr=0, count=0;
while(ptr!=a.size()){
if(a[ptr]==b[ptr]){
ptr++;
continue;
}
ptr++;
count++;
}
return count==1;
}
int ladderLength(string beginWord, string endWord, vector& wordList) {
wordList.insert(wordList.begin(), beginWord);
int n = wordList.size();
vector adj[n];
//Create undirected graph
for(int i=0; i
The graph creation will take time = (N^2)*M (where N is length of words array & M is length of longest word).
thanks striver for this content . i was able to solve this problem by own ,because the way you clear every concept
Understood! What an excellent explanation as always, thank you very much!!
Understood , a big fan of your zero typing errors !!!!! awesome video as always thank you sir ji
Thank you so much Striver for explaining problems in a very understandable way. Kudos to you for achieving lot at a very young age in the world of Top-tech
I remember this problem came in Leetcode daily in Feb 2022. Just copied the code and did not understand anything. One year later I understood the code. Thank you for the explanation.
excellent memory brother
You gave me the confidence of solving a leetcode hard! The step by step explanation is wonderful!!
Bravo, amazing!!
I was always afraid of this question. But you explained it sooooo well. Thanks a lot striver.
Yeah, it came out to be very simple.
It took me around one day to solve these question using my method but in the end it got TLE on leetcode and now after seeing the solution the questions seems easy.
Happy Teacher's day to you. You have helped so many of us to get their DSA concepts right. Kudos to you and thanks a lot :)
Hey Striver !! Your Videos are reaallly really helpful, easy to understand, and hats off for your amazing work and efforts.
Hey @striver this solution is exceeding time limit in leetcode, so instead of assigning all letter from a-z , I tried with letters only in wordList and it worked.
I just added below function to get all the letters into vector-
private:
vector letters(vector&wordList){
unordered_set st;
vector result;
for(auto it: wordList){
for(int i=0;i
I wrote the same code and it didn't give TLE.
Understood. You are brilliant in finding the suitable example to explain the algorithm....!! Thanks a lot🙌❤
My approach is different: This can be visualised as an undirected weighted graph with unit weights. We can created graph with word as each node and all neighbours are the words with only one character changed. Then apply exact same BFS as we implement in the previous video. No code changes at all. Quite simple and passes all testcases.
bool isAdjacent(const string& word1, const string& word2) {
int diff = 0;
for (int i = 0; i < word1.size(); ++i) {
if (word1[i] != word2[i]) {
diff++;
if (diff > 1) return false;
}
}
return diff == 1;
}
// Function to create the adjacency list graph
unordered_map buildGraph(vector& wordList) {
unordered_map graph;
// Add an empty list for each word
for (const string& word : wordList) {
graph[word] = {};
}
// Connect each word with its adjacent words
for (int i = 0; i < wordList.size(); ++i) {
for (int j = i + 1; j < wordList.size(); ++j) {
if (isAdjacent(wordList[i], wordList[j])) {
graph[wordList[i]].push_back(wordList[j]);
graph[wordList[j]].push_back(wordList[i]);
}
}
}
return graph;
}
int wordLadderLength(string startWord, string targetWord, vector& wordList) {
// If targetWord is not in the wordList, no valid transformation is possible
if (find(wordList.begin(), wordList.end(), targetWord) == wordList.end()) {
return 0;
}
// Include the startWord in the word list if it's not already there
if (find(wordList.begin(), wordList.end(), startWord) == wordList.end()) {
wordList.push_back(startWord);
}
// Build the graph from the word list
unordered_map graph = buildGraph(wordList);
// Perform BFS to find the shortest path
unordered_map distance;
for (const string& word : wordList) {
distance[word] = INT_MAX; // Initialize distances with infinity
}
distance[startWord] = 1;
queue q;
q.push(startWord);
while (!q.empty()) {
string currentWord = q.front();
q.pop();
// Traverse all neighbors of the current word
for (const string& neighbor : graph[currentWord]) {
if (distance[neighbor] == INT_MAX) { // If not visited
distance[neighbor] = distance[currentWord] + 1;
q.push(neighbor);
// If we reach the target word, return the distance
if (neighbor == targetWord) {
return distance[neighbor];
}
}
}
}
// If BFS completes without finding the targetWord, return 0
return 0;
}
can u share the code?
love it, just higher time complexity while building the graph--can go up to N^2
@@divyareddy7622 yeah
Thanks striver, Was able to solve it by myself although my approach was different from this,
I considered every node has a edge to every other node and each edge has a weight equal to the difference b/w the 2 vertices(words)
Then just used the shortest distance for an undirected graph using BFS algo
By node you mean the words right
@@arjunrai8126 yes
@@rajatsaraf237 by difference bw the words you mean the number of positions where the chars are different , right? Great solution
@@arjunrai8126 yes, that char difference can be considered as the weight b/w two words
W bro
you can make any difficult question way too easy. Just loved your teaching style .😊😊😊😊
Since its bfs, steps are going to be common for each level traversal. So no need to save the steps inside queue. Can maintain it outside & keep incrementing as we finish one traversal.
Your explanation is brilliant, bro!!
we can create graph from the word list and do topological sort on it. that will give only limited characters that exists in word list and we can try out replacing characters present in word list only. with topological sort the we get array as h,d,l,c,o,t,g.
Now I know before o I have 4 characters to try for first place i.e. h,d,l,c. then for g I have 2 characters o and t
with same BFS algoritm we can avoid trying out all a-z combination and also it makes sense logically because after transformation if word should be from word list then try out with characters that are only in wordlist. if z is not present in wordlist in any word why to try out with that. Considering if there is no cycle in it. This needs to be asked to interviewer whether there will be cycle in the graph. If cycle exists then the other approach is to iterate through all the wordlist and put them in set. Atleast you will be comparing less characters then a-z
Understood, quality content!
Simple bruteForce can be : for each element(say "hit") in list, compare it with the all other elements. If the only 1 char varies that we can add as part of adjacency List of "hit" and so on.
Very clear explanation. Hats off to your effort .
Was able to solve the problem on my own only because of you...Thank U😊😊
Understood Bhaiya..
Such a great explanation! Please continue the playlist
I did like and kya hi samjhate ho yrr bhaiya, maza hi aa gaya. UNDERSTOOD
your explanation made it too easy🤗
excellent explanation as always, thank you very much!! Amazing playlist
Thank You So Much for this wonderful video..........🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
For string set.find() method takes O(N) instead of logn.... It increases the complexity
thanks striver bhai, for this awesome series
Thanks a lot, Striver. Best explanation . 😀
Briliiantly explained
When I read the question for the first time, I was so scared and I had no clue how to proceed.
I was thing of sets and recursion but I did not even try coz Iwas scared.
But u made it soo simple that after watching half of the video, I did it in 1 go .
U r amazing striver.
please continue to bring such playlists , its so helpful
were you thinking about backtracking too in first go?
Understood!! Very well explained!!!
understood. Able to write the code by myself after getting the intuition.
Best series on graph
Great work, Raj!
Understood 🎉
understood! amazing like always.
understood....your explainations are lucid ....keep up the good work...
this question was amazing , thanks striver
GOAT❤ you are great striver
great explanation!!
One can figure out that it was a graph question by looking at src & dest strings, & shortest sequence of transformations were asked...So, some 'feeling' of shortest path from src (beginWord) to dest (endWord) comes...
UNDERSTOOD striver!
yours HArdwork is really ❤
thanks striver for such wonderful explanation.
best method to solve this question 😁thanks striver
I got your algo and tried myself , just make a little mistake but it was corrected bcoz of your code
Thank you for such a wonderful Explanation !!!!
I have a different solution, i created a graph representation using adjacency list (word is the key, words it can transform into are list values), then applied shortest distance method using dijkstra's algorithm.
can u please share your code....im having difficulty in making adjacency list
14:36 Odissshhhaa
another approach is to use bi-directional BFS. same time complexity but half the runtime
understood, thanks for the detail explanation
Hi, awesome video. Just add one base condition to handle this case, according to your code it will give 1 but it should return 0..
1
der
der
der
It's a high time youtube should add a super like button.
As always AMAZING!
Instead of converting str to all the possible strings
we can simply compare str with all the strings in word_List and see if the difference btw them is 1 or less and if so then add in queue to do BFS.
understood ❤
Amazing explanation and logic! kuch bhi bolo, Striver bhai is built different!
thnaku striver ...you are making my carrir
Striver, thank you so much for graph series, per please can you also make series on segement trees, tried learning from so many places, but nowhere satisfactory content is found, bana do na ek series please
How u make the hard question like a cake work amazing brother ......
once i knew it was a graph question, i tried this myself :
class Solution {
private:
bool check(const string& w1, const string& w2) {
int differences = 0;
for (int i = 0; i < w1.length(); ++i) {
if (w1[i] != w2[i]) {
differences++;
if (differences > 1) return false;
}
}
return true;
}
public:
int ladderLength(string beginWord, string endWord, vector& wordList) {
bool possible = false;
int index = -1;
for (int i = 0; i < wordList.size(); i++) {
if (wordList[i] == endWord) {
index = i;
possible = true;
break;
}
}
if (!possible) return 0;
wordList.insert(wordList.begin(), beginWord);
int n = wordList.size();
vector adj(n);
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (check(wordList[i], wordList[j])) {
adj[i].push_back(j);
adj[j].push_back(i);
}
}
}
vector distance(n, INT_MAX);
queue q;
q.push({0, 0});
distance[0] = 0;
while (!q.empty()) {
int node = q.front().first;
int dist = q.front().second;
q.pop();
for (int val : adj[node]) {
if (distance[val] > dist + 1) {
distance[val] = dist + 1;
q.push({val, dist + 1});
}
}
}
if (distance[index + 1] < INT_MAX) return distance[index + 1] + 1;
return 0;
}
};
Striver you are the best
A Small Optimization Can be MADE further ...
So currently, for each index of the string we are traversing 26 times representing 26 characters / alphabets in the English Dictionary. But what if we exactly store the possible set of characters that are only available in wordList.
There comes the optimization where we store set of all alphabets for each index & always iterate those in the while loop of queue.
Meeri jhann!! understood!
really don't understand why this is marked as "hard", this is one of the easiest problems, and it can also be considered as tree too.. regardless.. thank you Striver❤
understood..thanks bhaiya
Hey Striver! Understood the concept thoroughly. Had one doubt, my first intuition was to use DP in this question instead of graph. SO why graph is more optimal then dp?
1) The question has a source to destination pattern, it's guiding you a path that you start from this word and end somewhere... This intends graph... Also they are saying "shortest" while giving you source and destination so you should think of shortest path algorithm
2) Reason why you thought of Recursion/DP is because we are trying every letter combination which is fair to think but you apply DP when you have a vast search space and you have no clue which can be the answer but here they've given you the dictionary list so they are telling you the search space
We converted DFS TO BFS😅😅
One different approach is make a graph with edges between hit and hot, hot and dot, hot and dot, hot and lot, .... And now you have to find the shortest distance between two points (beginWord and endWord) using BFS.
Yeah
Understood sir thankyou sir❤🙏🙇♂️
Nicely Explained
Kudos to you sir!!
At instead of set i have used map O(1) in every case : )
there can be collisions in a hashmap
is use of unorderedmap better of unordered set and why??
amazing explanation bhai 🤩🤩🤩🤩
Thanks Strivers!!!!
very nice explanation
Understood bhaiya
understood striver 🙌
fabulous striverrr
Great explaination
Understood Sir!
Understood! 🤩❤🔥
Thank you sir
Understood 👍🏻
Understood,sir.
understood sir🙇♂🙇♂🙇♂
UNDERSTOOD.
Understood 😇
understood, Thank you.