Daily
https://leetcode.com/problems/reordered-power-of-2/description/?envType=daily-question&envId=2025-08-10
Hint:check all power of 2, try to think how sorting might help here
#daily
https://leetcode.com/problems/reordered-power-of-2/description/?envType=daily-question&envId=2025-08-10
Hint:
#daily
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Streaming today is 50/50. It will be shifted in case I cannot make it today's evening.
Topic - I want to cover sharding topic fully. So next streams for system design section we are reviewing how sharding works in different storages, and gonna implement POCs for each of the type🔥🔥🔥
Topic - I want to cover sharding topic fully. So next streams for system design section we are reviewing how sharding works in different storages, and gonna implement POCs for each of the type🔥🔥🔥
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I guess recording will be there , played a lot of games, not on diamond, but on plat level. But yea, we did stream, and it was not lagging that much
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Solutions and hints:
hint1:to get the all powers of 2 that number consists of - you basically need to get it's binary represantation as this is how binary number works {for each index I in binary representation, the number can be got by SUM( VALUE[I] * 2 ^ I )}
for binary 1011 we get => (1 * 2 ^ 3) (0 * 2 ^ 2) + (1 * 2 ^ 1) + (1 * 2 ^ 0) = 8 + 0 + 2 + 1 = 11.
So first hint is how to get your array - just decompose it's binary representation
hint2:think about how large the array of your num can be, given 1 <= n <= 10^9. Recall 10^ 9 comparing to 2^32, and apply bruteforce.
hint1:
for binary 1011 we get => (1 * 2 ^ 3) (0 * 2 ^ 2) + (1 * 2 ^ 1) + (1 * 2 ^ 0) = 8 + 0 + 2 + 1 = 11.
So first hint is how to get your array - just decompose it's binary representation
hint2:
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Decided to take a nap in the office for fun, never did it
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I guess you might remember demo for Doom 1 or something like that, where AI was generating the next frame?
My friend just shared the Minecraft AI version
https://youtube.com/shorts/ZbaxA5r4k2k?si=pg34NcG4tZU3p6QI
My friend just shared the Minecraft AI version
https://youtube.com/shorts/ZbaxA5r4k2k?si=pg34NcG4tZU3p6QI
YouTube
Digging down in AI Minecraft... #minecraft #ai #aiminecraft
sub if u want.All Socials:Solo.to/zyphron_
Daily.
Today is DP!!
https://leetcode.com/problems/ways-to-express-an-integer-as-sum-of-powers/description/?envType=daily-question&envId=2025-08-12
Hint:uniqueness is ensured same way like when you are generating the combinations in this problem: https://leetcode.com/problems/combinations/description/
#daily
Today is DP!!
https://leetcode.com/problems/ways-to-express-an-integer-as-sum-of-powers/description/?envType=daily-question&envId=2025-08-12
Hint:
#daily
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I wanted to discuss one thing, about time complexity analysis
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Do you see the difference between these 2 DP functions?
I'm a bit БОМБЛЮ because, I have seen a lot of similar problem analysis when authors would name both of these as O(2^n).
FUCK NO. these are different complexities, and today's daily proves it, you can submit it as comb2 and you'll get Time Limit Exceeded.
Without memoization the complexities:
comb time complexity O(2^n), as 2 decisions only
comb2 time complexity is approximately O(n/2 * 2^n) => O(n* 2^n), as on every next call we are doing n - k calls. It will approximate to n / 2, a bit hard to analyse but definitely not pure 2^n
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Do you see the difference between these 2 DP functions?
I'm a bit БОМБЛЮ because, I have seen a lot of similar problem analysis when authors would name both of these as O(2^n).
FUCK NO. these are different complexities, and today's daily proves it, you can submit it as comb2 and you'll get Time Limit Exceeded.
Without memoization the complexities:
comb time complexity O(2^n), as 2 decisions only
comb2 time complexity is approximately O(n/2 * 2^n) => O(n* 2^n), as on every next call we are doing n - k calls. It will approximate to n / 2, a bit hard to analyse but definitely not pure 2^n
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Today's daily is chill one, easy:
Brute force will fit.
https://leetcode.com/problems/power-of-three/?envType=daily-question&envId=2025-08-13
#daily
Brute force will fit.
https://leetcode.com/problems/power-of-three/?envType=daily-question&envId=2025-08-13
#daily
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Follow up in constant time O(1).
We take the maximum possible power of 3 num in the constraints `n <= 2^31 - 1`.
And then if number is positive - check if the reminder will be zero
If you have only 1 run, then, yeah, you might calculate all power of 3, put them to map, and then in constant time chech if n is in map
And then if number is positive - check if the reminder will be zero
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