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I'm afraid of this problem because of it's name "Kadane's Algorithm" ... After ur video I went to GeeksForGeeks and solve this also solved it in Leetcode... Thankyou so much!!... I thought DSA is very hard for me..But ur explanations made it very easyy!!

Great Explanation !!!!

The logic sounds very simple but its difficult to understand for cases where the subarray does not contain the starting (leftmost) or the ending (rightmost) element. When the sub array is in between the array

Hi, great explation you keep up to the point, but I believe you can go more in depth while using the loops for, explaining how the second for loop work with some interactive example. Great work thanks

Thank you so much for explaining each line in a very simple manner.

most of the you tuber have taken the same example and are doing the same thing but in question we have to compare NUMS1 and NUMS2 the find it

Literally the best one

Thank you so much! First time seeing someone explaining solution as good as NeetCode1

brilliant🤩

guys who ever is confused with '^' in Stringbuilder it is to handle below kind of tree node TreeNode root = new TreeNode(12); TreeNode subRoot = new TreeNode(2); where 12 (tweleve) will be treated as 1,2 subtree as 2 it gives true to figure out item or data in node we are using ^ as seperator still if u need confirmation pls refer test case created by nikhil in description code and nikhil small suggestion pls explain or stick to same logic during dry run , this will give more clarification and reduces doubts , in last example no null were included in ur explanation rest vedio is very helpful thank you ❤

very nice and clear explanation.

❤

bhai [1,1,1,1,2,2,3,3,4] and k = 2, test case hi galat hai kyuki answer unique nahi hai, it is clearly mentioned in constraints, It is guaranteed that the answer is unique. toh 2 and 3 ki freq same nahi ho sakti aur agar hogi toh k ki value 3 hogi.

You have a god gifted talent of explaining things in a simple manner, this is sign of a true teacher, thank you so much 🙏.

IF C++ CODE IS FAILING this is because substring function it shoule be like s.substr(i-1,1) and s.substr(i-2,2) the second parameter is the length of the string. this should work class Solution { public: int numDecodings(string s) { int n = s.length(); vector<int> dp(n + 1); dp[0] = 1; dp[1] = s[0] == '0' ? 0: 1; for(int i = 2; i < n + 1; i++){ int d1 = stoi(s.substr(i-1,1)); int d2 = stoi(s.substr(i-2,2)); if(d1 >= 1){ dp[i] += dp[i-1]; } if(d2 >= 10 && d2 <= 26){ dp[i] += dp[i-2]; } } return dp[n]; } };

loved this really loved this!!!

isme ek problem hai duplicate triplets ko lekr ... triplets double print horhe

So, during insertion into the priority queue, it takes nlogn and also during polling it will also take nlogn. Is my understanding correct?

Also for negative numbers how to make it work in python

Gr8, thanks, could you please explain why cumulative and placing in output array with -1, should work

tried watching but you just explained what the code does and its too confusing for me to undertsand what for each freq [c -'a']++; Does

for cpp : class Solution { public: int sumOfUnique(vector<int>& nums) { unordered_map<int,int>mp; for(int i:nums) mp[i]++; int sum=0; for(auto it:mp){ if(it.second==1) sum+=it.first; } return sum; } };

Impressive

thank you bro

Awesome explanation sir

new subscriber

Nikhil, your code would not work for the test case you mentioned: [1,1,1,1,2,2,3,3,4] & k=2 This code is getting submitted on Leetcode because there it is mentioned that unique answers only. But in the about test case: We should get [1,2,3] as ans for k=2. You cannot assume the res array of size k since there might be duplicacy. Otherwise solution works fine for the Leetcode problem. Here is the Code which will cover duplicacy as well. class Solution { public static int[] topKFrequent(int[] nums, int k) { int n = nums.length; List<Integer>[] bucket = new ArrayList[n + 1]; HashMap<Integer, Integer> frequencyMap = new HashMap<>(); ArrayList<Integer> resultList = new ArrayList<>(); for (int num : nums) { frequencyMap.put(num, frequencyMap.getOrDefault(num, 0) + 1); } for (int i = 0; i <= n; i++) { bucket[i] = new ArrayList<>(); } frequencyMap.forEach( (element, frequency) -> { bucket[frequency].add(element); }); for (int i = n; i >= 0; i--) { if (bucket[i] != null) { resultList.addAll(bucket[i]); if (resultList.size() >= k) { break; } } } int[] result = new int[resultList.size()]; for (int i = 0; i < result.length; i++) { result[i] = resultList.get(i); } return result; }

explanation was amazing

🤩

ur explanation is amazing..Thankyou sm!!!

Before watching this video, watch Counting Sort, its a prerequisite.

This solution is giving "time limit exceeded" error n leetcode...

Bhai start dsa course atleast for members, ill buy a membership for sure.

This solution has runtime of 10ms and beats only 68%. Do you have a faster solution?

Sir I've a better approach for this We can set the minimum value to the (maximum value/no of days) Because the range can better shorten by this logic.

awesome solution buddy!!

Definitely leetcode tagged this wring it is a medium level problem

Sir u ar doing work plz keep doing it... I will promote your channel... U got great content truly helpful ❤❤❤❤

This is just a fibonacci series answer.

how to get that memozation logic?? like how will I know the rules of filling the table and analysing the answer??

Good work sir 👏

I'm confused, Why do we mark a node as visited as soon as we add it into the queue as oppose to while we are actually polling it in the code? in DFS we only marked a node as visited when we popped it from the stack.

Nice explanation sir

As of this writing LeetCode added a new test case which cause the above code to fail. The logic is still correct though. The test case fails because of overflow error. With just a small fix, the code still works. Have a look at the updated code on my Github link :)

Soooo I have watched this video multiple times and have looked at the question you show in the beginning of the video. Though I understand what you’re doing with using an index to identify duplicates what I don’t understand how this would work based off your question without knowing the range of numbers in the array. Your example uses something like this: [2,4,1,6,3,1]. How will this solution work with an array like this based off the question?[3,16,1,4,2,1].

This is work ❤❤

Girl did 2 sum and 3 sum today!! Keep going, faang is waiting for me

amazing

The solution is wrong. Also why did you convert all the remaining elements to zero in the last step? If you take the example from striver's video he particularly said about that element.

i did not understand fast=nums[nums[fast]] .if nums[fast] is 3?