there's a small mistake in the dry run in moore's algo at last votes for majority 2 is 1,when majority reaches 3 the votes will be 0, so, majority will be updated in the next iteration so majority will be update as 1, please check, explanation is too good!!!
the dry-run is mainly to understand the simplicity of the code. The exact working code is available in the description too. Mostly you should understand the approach and how you are solving the problem :)
Great video man! Loved this explanation, you are highly underrated. If it's possible can you do python solutions as well for future leetcode videos? Thanks again!
i would advice you to focus on the problem solving method, rather than the language. Trust me...languages will come and go...get your basics right first :)
the voting method returns 1 for me for array {1, 2, 2, 2, 3, 3, 1 } so is that logic correct? I think after your logic, we need to check again in the array if the count of the majority element is greater than (n/2) to be considered as the majority. In my case, the majority is returned as 1 but 1 is repeated 2 times which is not greater than the expected majority which is (>3). here we can suspect 2 could be the majority but it's not because it is not repeated more than 3 times.
Majority element means the element which occurs more than n/2 times. Your test case is invalid, as it does not have a majority element. What you are talking about is the element occurring maximum number of times.
@@foodandjournieswithme8788 check your test case and code again. Here is the verified output: github.com/nikoo28/java-solutions/blob/master/src/test/java/leetcode/easy/MajorityElementTest.java
The mistake in the code is that the loop starts with i = 0, which causes the initial element to be counted twice. Specifically, when the loop starts, nums[0] is already assigned to the majority and the vote is set to 1. The loop then starts from i = 0, incrementing votes for the same element. The correct approach is to start the loop from i = 1. for (int i = 1; i < nums.length; i++)
@@abhiguptamusic That condition is absolutely correct, whenever votes become =0 you need to update your majority candidate, and then increase the vote count. What error do you see in the condition? Did you try running the code?
Slick & straight! Thank you for breaking the complexity.
You the one of the best here on TH-cam, the way you teach, the way you keep every in a structured manner is super commendable. Subscribed.
thanks for the kind words
Keep making these videos bro, the quality is Amazing!
Bro, the quality of your content is exceptional. You deserve more subscribers. Thank you brother.
I wish that too 😄
there's a small mistake in the dry run in moore's algo at last votes for majority 2 is 1,when majority reaches 3 the votes will be 0, so, majority will be updated in the next iteration so majority will be update as 1, please check, explanation is too good!!!
the dry-run is mainly to understand the simplicity of the code. The exact working code is available in the description too. Mostly you should understand the approach and how you are solving the problem :)
Thanks for this great explanation
Amazing exaplanation
You've explained in a very simple way!!
pls make a video on peak element ... your videos are so helpful!!!!
After watching 5 videos finally understood ... kaafi acchaa explanation...loved it
Thanks for liking
great solution as always. Thanks alot bhaiya please keep on making such amazing videos.
So nice of you
Beautifully Explained
Your teaching is superb. You have a new subscriber.
Thanks a lot 😊
Next level & Awesome explaination with cutest smile. Thank you😊
that is so sweet of you
great explanation
this was very helpful 😀 Thank you
Superb Explanation!
Thank you 🙂
Thanks a lot sir ! Really helpful
it's tough to get an optimized solution 😔.... but I will try to reach it on my own 😊.
Awesome explanation 😮😮
SO CLASSY AND please make the playlist of data structures important questions too
playlist: th-cam.com/play/PLFdAYMIVJQHM8Kh5i8P2lGIbJXFPBelRI.html
thank you, your dry run really helped
Thank you. Very good explanation!
Great video man! Loved this explanation, you are highly underrated. If it's possible can you do python solutions as well for future leetcode videos? Thanks again!
i would advice you to focus on the problem solving method, rather than the language. Trust me...languages will come and go...get your basics right first :)
Glad i found your channel
better explanation than Striver .
the second soln was impressive,
super sir good explanation
Amazing!
amazing amazing!!!!!
Thanks
Amazing great explanation
Why there are only three types of numbers? In array
you can have as many types.
Great video, you explained it so well.
Thanks !!
Very Well Explained
Thank you so much 🙂
perfect teacher
perfect student :)
LEGEND 🖤
*Sir Please Complete top 150 interview Questions First from Leetcode 🔥*
there are some problems from that list that I have covered...adding new solutions every week :)
@@nikoo28 *Thankyou Sir Loved your teaching Very Clear & Upto the point*
@@nikoo28 thank you!
great video
Thanks!
You're the best👍💯
You are!
Thanks man, good explaining, if i land a job, ill send you some money 😁
haha..thanks a bunch
the voting method returns 1 for me for array {1, 2, 2, 2, 3, 3, 1 } so is that logic correct?
I think after your logic, we need to check again in the array if the count of the majority element is greater than (n/2) to be considered as the majority.
In my case, the majority is returned as 1 but 1 is repeated 2 times which is not greater than the expected majority which is (>3). here we can suspect 2 could be the majority but it's not because it is not repeated more than 3 times.
Majority element means the element which occurs more than n/2 times. Your test case is invalid, as it does not have a majority element.
What you are talking about is the element occurring maximum number of times.
Can we do using 2 pointer
Give me a pseudo code for your approach
What if there is no majority element? How to handle that?
then it will be an entirely different problem. What are you looking to find?
It gives error when nums=[6,5,5]
What error are you getting? I tried the case again and it gives 5 as the output
@@nikoo28 sir i got output as 6 in the same code
@@foodandjournieswithme8788 check your test case and code again. Here is the verified output: github.com/nikoo28/java-solutions/blob/master/src/test/java/leetcode/easy/MajorityElementTest.java
The mistake in the code is that the loop starts with i = 0, which causes the initial element to be counted twice. Specifically, when the loop starts, nums[0] is already assigned to the majority and the vote is set to 1. The loop then starts from i = 0, incrementing votes for the same element. The correct approach is to start the loop from i = 1.
for (int i = 1; i < nums.length; i++)
Can we get n/3 solution as well?
do you have a link to the problem?
Wrong Code (Wrong understanding of mine)
What part do you think is wrong?
@@nikoo28 first condition vote==0
@@abhiguptamusic That condition is absolutely correct, whenever votes become =0 you need to update your majority candidate, and then increase the vote count. What error do you see in the condition? Did you try running the code?
@@nikoo28 the code is not working with another test cases
@@nikoo28 please try with these test case [1,1,2,3,4]