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To anyone who is still finding it difficult to understand, following might help. The idea is to keep track of max profit by end of each day (ith day) when you have completed one or more of following transactions so far (by the end of ith day): 1. You buy for the first time: fb -> profit after first buy (first buy up until the ith day). 2. You sell for the first time: fs -> profit after first sell ((first sell up until the ith day)). 3. You buy for the second time: sb -> proft after second buy (second buy up until the ith day). 4. You sell for the second time: ss -> profit after second sell (second sell up until the ith day). The profit after first buy (fb) up until the ith day will be the max of i) The profit if you buy for the first time on the ith day i.e., -prices[i]. ii) The fb profit seen so far up until the (i-1)th day (you are keeping a running max, so one variable(fb) is enough to keep track of it) The profit after first sell (fs) up until the ith day will be the max of i) profit if you sell for the first time on ith day. This will also depend on the profit due to first buy up until the ith day. So fs = fb + prices[i] ii) the fs profit seen so far up until the (i-1)th day (you are keeping a running max, using fs variable only) Similarly, the profits for sb and ss would depend on fs and sb respectively. Hope it helps. Please don't hesitate to reply if you have any questions. :)

Woww What a video!! I really liked your simple solution and explanation.I saw many youtubers video of this same question including Striver and more but you just explaiend the solution simply in 11 minutes.

You are so beautiful 😍😍 Are you single dear?😍😊

recursion hi likh deti !!

sundar thi issi liye aa gya video dekhne waise mujhe java aati bhi nhi hai phir bhi pura dekh liya ❤😊

The fact that you don't follow a step by step manner from naive to optimal shows your incompetence (at least in teaching).

Very impressive solution:)

Thanks 👍

You may need to use digital pen for better explanation just a suggestion

Amazing Way to Teach, Mam, pls provide slide u shared at 1:23, it would be very helpful.

Why not just do the number to string conversion in the sort method? That will avoid the first loop over the entire number list. 🤔

this method is not intuitive...dp solution is more intuitive and then convert that solution to sapce optimization

How to find and return the indices of best day to buy and best day to sell with max profit?

great...

for a newly added test case. Plz use(int temp) as( long temp) and in helper function (int sum) as (long sum).

In the first example, given that the root value is 10 and target sum is 8, how is it possible that we have 3 results? If we travel downwards and keep track of the path, then sum will never be 8 (target sum) simply because 10 (root value) is greater than 8 (target sum). The only solution I can think of is summing up the path upwards, which goes against the description: Quote from description: The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes). What am I missing here?

Mam please explain more questions.on graph ,linkedlist ,stack and dp which asked in mid and top level companies in java

Thank u so much. I didn't take a look at you code just I saw the commented instructions and solved the problem. Not always to think recursively.

And we were fools to use DP solutions. This solution is best

Can we use 2 pointers instead of a priority queue in the sorted intervals, ( we can maintain a counter which keeps on incrementing as long as we are having clashed meetings consecutively, else re initialize it with 1 while iterating in the intervals array )

Your code passed all the test cases on very first trial you are a genius

We can solve all variations of stock buy sell with this 1 code. ruclips.net/video/G_BICqMdoYU/видео.html

We can solve all variations of stock buy sell with this 1 code. ruclips.net/video/G_BICqMdoYU/видео.html

We can solve all variations of stock buy sell with this 1 code. ruclips.net/video/G_BICqMdoYU/видео.html

We can solve all variations of stock buy sell with this 1 code. ruclips.net/video/G_BICqMdoYU/видео.html

This is Even Best for C++ users I modified it Little Bit class Solution { public: int minMeetingRoom(vector<vector<int>>meetingTime) { sort(meetingTime.begin(),meetingTime.end()); priority_queue<int,vector<int>,greater<int>> pq; pq.push(meetingTime[0][1]); for(int i=1;i<meetingTime.size();i++) { if(meetingTime[i][0] >= pq.top()) { pq.pop(); } pq.push(meetingTime[i][1]); } return pq.size(); } };

full confuse🙂

You are the best till now.

Awful!

why do we have to sort on the basis of start time ?

Why Summation of dp[i-1] and dp[i-2] that I cannot understand.

why have to put +1 when we return max of L,R?

A simpler version could be: public int eraseOverlapIntervals(int[][] intervals) { Arrays.sort(intervals,(a,b) -> a[0]-b[0]); int end=intervals[0][1], res=0; for(int i=1;i<intervals.length;i++) if(intervals[i][0]<end){ end=Math.min(end,intervals[i][1]); res++; }else end=intervals[i][1]; return res; }

A simpler version with the same approach can be: public int minMeetingRooms(List<Interval> intervals) { if(intervals.size()<=1) return intervals.size(); PriorityQueue<Integer> pq = new PriorityQueue<>(); intervals.sort((a,b) -> Integer.compare(a.start,b.start)); pq.add(intervals.get(0).end); for(int i=1;i<intervals.size();i++){ if(pq.peek() <= intervals.get(i).start) pq.poll(); pq.add(intervals.get(i).end); } return pq.size(); }

So beautiful the teacher 😉

why do you need the has?

wooow officially the worst explanation on RUclips....

Can someone post solution in cpp with the same logic

thank you so much for teaching us this problem. A recruiter that does not like me will ask why is it so slow? why is 90% of other submissions faster? I'll resaearch it! is there a book besides cracking the interview? Your so kind for putting a solution to this problem that is common to asked in an interview

noice

Thank you ✨

implementation in ruby ``` def add_two_numbers(l1, l2) pointer_to_head = ListNode.new # dummy tail = pointer_to_head local_sum = 0 until l1.nil? && l2.nil? && local_sum == 0 if l1 local_sum += l1.val l1 = l1.next end if l2 local_sum += l2.val l2 = l2.next end tail.next = ListNode.new local_sum % 10 # hook up local_sum = local_sum / 10 # serves as a carryover tail = tail.next # reassign running pointer to the newly created node end pointer_to_head.next end ```

guys, marry this type of women.

Thank You !

thhoda hindi me hi bol deti didi...ye itna jyada technical word kahogi to kaha se smjh aaega ....ham log khud yaha seekhne aae hain.

The easiest and fast solution that i found on youtube thank you very much!

Why don't you explain the problem in notepad or graphical form. It would be helpful for beginer like us instead just speaking and explaining. ??

5/5

without using pen and paper she explained it and I can write it easily.

amazing. calm and simple explanation. no over confidence. better than DSA gurus.