Time limit exceeded

dwarfed the problem🤔

it too bad explanation

Well explained, but I think the code can be reduced and improved

Nice work. You can optimise a little bit by removing redundant code. just check "total == diff" once and perform the "while total == diff" until you get the best j and after you break you can check only once if(a[i].first-a[j].first+1<mini) inside the "total == diff" case and that would work as well. Good job though helped me get the intuition for this question.

worst naming of variable i have ever seen

kya bol raha hain khud samajh raha hain kya? English bola zaruri hain??

bad explanation

Please correct problem number

Regarding Leaping task, while i was watching for first time it was really frustrating for me but after a quick 2nd recap it all made sense

Great video I think we can use stack to get rid of using timestamps

best one by far...Can you explain why did you take 1e11 in mini?

yeh bkl ytuber bnega haha rxxndva

great

Brilliant solution, explanation and code!

u know what the most comedy problem in seen in leetcode yarr ,bcs code logic is already return in question😅

useless video. i have already sovled this problem with this O(n*m) complexity came on youtube for watch a better solution . if you wanted to make the vedio solution then it should have been a better solution of O(n+m) time even a child can solve O(n * m). just don't upload videos on youtube just for shake of it. time waste

wooow now i get it fot the first time after watching so many lectures

best explanation found!

great explanation

tq 😘

Perfect :)

Thanks for the awesome explanation

Hi, Juraj here, one of the maintainers of Cyclops 👋, Loved the video, great to see it from the perspective of a contributor!

Bhai difficult concept h hindi m pdhaya kro

Excellent Thank you so muchhhhhhhh

very good explanation bhai even i can't solve it

in second question why did you take that condition, like that condition will never come right as we are eliminating all the odd n so a can never be even

It's also Pirates of carribbean

Bhai , everyone does cp in c++. I am learing dsa in java. How to do cp in java? or should i swtich to c++.? Can you give resources how to start cp in java, or should i switch to cpp

nice solution man. loved it

What I love here is that he is focused on the problem while his family is just going about their daily routine as if he isn't doing anything. hehe

GREAT EXPLANATION ❤

Insightful.................

hello brother, I'm new in coding era... and your idea in this video helps me a lot to understand this logic... after this made little bit easy code thn you.. here it is int n, arr[123]; cin >> n; for(int i = 1; i <= n; i++){ int x; cin >> x; arr[x] = i; } for(int i = 1; i <= n; i++){ cout << arr[i] << " "; } return 0; btw thanks a lot

Thanks bro without watching your full video I got direct hint from your code and solved it by myself.

Quite an ingenious smart trick. i was thinking of actually tracking the addend throught the array. 1 + 2 + 4 + 8 and so on... Awsm implementation

Thank you sir ❤

I came with the same approach but unable to write the code tbh

bro group ur video in playlist u will get more views js 30 day challenge will need its own playlist

it makes sense but the way of explanation is poor

linear traversal should not be n*logN.

I don't think dp[i/j] + j means pasting i/j A's j times.......it means copying 1 time and pasting j-1 times making it a total of 1+j-1 = j operations. furthermore, I think j doesn't need to iterate from 2 to i ...iterating from 2 to i//2 'll suffice My C++ solution: class Solution { public: int minSteps(int n) { vector<int> dp(n+1,1000); dp[1] = 0; int count; for(int i = 1;i<=n/2;i++){ count = 2 + dp[i];//start with 1 copy+1 paste & then increment paste by 1 for (int j = 2*i;j<=n;j+=i) dp[j] = min(dp[j],count++); } return dp.back(); } };

Bro . Stop saying COOL. again and again 😑😑😑

i know this is a easy one, but still I was missing something and that's what your vdo helped me to see thanks :)

mate don't make videos in english bcz you can't right words to explain something and it messes up the entire explaination

no wonder u have 89 dislikes

please always provide java code link also

but the k is atmax 10 so we will have atmax 20 elements in the union so the time complexity wouldn't be O(N*20) i.e O(N) confused

NICE SUPER EXCELLENT MOTIVATED