Hi Aryan! Amazing explanation indeed. Would it be possible for you to share the PDFs of your explanation you used in the video. It would be really great to glance over after watching the video. Thanks.
Thanks brother ! I spent whole day trying to figure out the logic of Hard problem... but after watching this everything is clear.... video was long but necessary.
@Aryan I wanted to share some feedback on your video. I noticed that there were a few instances where the same sentence was repeated multiple times. I believe the video could be even more engaging and concise without those repetitions. Just a suggestion for future videos to keep the content as clear and streamlined as possible. Thank you for your hard work!
in java bitset class cannot be used with "bitwise and" operator. also the inbuilt "and" function stores the result of "and" to the first operand and returns void. Also there is no way to left shift except shifting each set bit left in a loop.
DP se nahi kar sakte kya ye problem i am trying to do with dp. 1st one is done by recursive dp int n; int solve(int i,int x,vector& v,vector& dp){ if(i==n) return x; if(dp[i][x]!=-1) return dp[i][x]; int nottake = solve(i+1,x,v,dp); int take = 0; if(x= 0 && x - v[i - 1] < v[i - 1]) { take = v[i - 1] + pre[x]; } cur[x] = max(take, nottake); } pre = cur; } return pre[0]; } it is showing wrong answer how to correct this please help me to solve this prolblem
You are always my last option to understand a problem, not because you are bad at teaching, rather it means no it's time that I need my best weapon against that question.
i was trying to do dp with binary search, because we require next Element to be greater than the previous one, i tried to do like this, jaise aapne ISS VIDEO mein bataya tha, bhaiya aap bata sakte ho ki iss question mein ye solution work kyo nahi kar raha. czcams.com/video/Uz2iNhjGGm0/video.htmlsi=Tg7B_P-OXmPaHGDT (edit) : int maxTotalReward(vector& nums) { // Remove duplicates and sort the nums vector sort(nums.begin(), nums.end()); nums.erase(unique(nums.begin(), nums.end()), nums.end()); int n = nums.size(); // Define the solve function with memoization function solve = [&](int t) -> int { if (t == 0) { return 0; } auto p = lower_bound(nums.begin(), nums.end(), t); if (p == nums.end()) { --p; } if (*p == t) { return t; } else if (*p > t) { if (p != nums.begin()) { --p; } else { return 0; } } int ans = 0; while (p >= nums.begin() && *p * 2 - 1 > ans) { ans = max(*p + solve(min(t - *p, *p - 1)), ans); if (p == nums.begin()) { break; } --p; } return ans; }; return solve(nums.back() * 2 - 1); } i found this binary search solution but not able to understand it, can anyone please tell why the call is from (nums.back() * 2 - 1); and not (nums.back() - 1); simply
I know the video is wayy beyond 20mins(promised time), but pls try to jump on respective Timestamps for understanding 🥺
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And haan iss baar 45k paar ❤
I think in First DP approach the ans should be divide by 2 before returning.
Hi Aryan! Amazing explanation indeed. Would it be possible for you to share the PDFs of your explanation you used in the video. It would be really great to glance over after watching the video. Thanks.
please provide with the pdfs especially for long videos
Thanks brother ! I spent whole day trying to figure out the logic of Hard problem... but after watching this everything is clear.... video was long but necessary.
Honestly was so lost when I saw bitset solutions because I've never heard of bitset, watching the video helped a lot
isse solve krte hue aaj pehle recursion m TLE a gya tha then memoization pe MLE then tabulation krte krte mene haar maan li thi
Gajab explanation ❤
Hats off to your hard work and patience
One of the finest explanation I've seen . Bhaiya bhot maza ayaa
Yaar maza aa gya. Lovely explanation.
Good Explaination.Thanks for the timestamps.
Ho jayega 45K subscribers
very nice bhaiyya.explanation give me the intuitions.
it was so good! thank you bhaiya!
Nice Explanations, Thank you sir !!
great job !!
@Aryan
I wanted to share some feedback on your video. I noticed that there were a few instances where the same sentence was repeated multiple times. I believe the video could be even more engaging and concise without those repetitions.
Just a suggestion for future videos to keep the content as clear and streamlined as possible.
Thank you for your hard work!
thanks good explanantion aryan
thanks
Great optimisation
Great content!
waiting for this video !!!!!!!!
Nice explanation
You're awesome.
Good Stuff, Thanks
Great video bro
in java bitset class cannot be used with "bitwise and" operator. also the inbuilt "and" function stores the result of "and" to the first operand and returns void. Also there is no way to left shift except shifting each set bit left in a loop.
Getting TLE with bitset. #justJAVAthings
They should just rename it as class ShitSet at this point.
Why taking a vector dp(n+1) giving tle? can anyone explain?
i also did same like you,
the tle is due to search operation in map which takes O(logn) extra to find the sum when compared to vector
How can u take the reward sum value atlast will be the max of rewvalues in the above sum, before 9 mins
DP se nahi kar sakte kya ye problem i am trying to do with dp. 1st one is done by recursive dp
int n;
int solve(int i,int x,vector& v,vector& dp){
if(i==n) return x;
if(dp[i][x]!=-1) return dp[i][x];
int nottake = solve(i+1,x,v,dp);
int take = 0;
if(x= 0 && x - v[i - 1] < v[i - 1]) {
take = v[i - 1] + pre[x];
}
cur[x] = max(take, nottake);
}
pre = cur;
}
return pre[0];
}
it is showing wrong answer how to correct this please help me to solve this prolblem
Any prerequisities for this video Except BIT MANIPULATION cause I have no knowledge on DP !! ?
22:26 can anyone please give me the mathematical reasoning behind reducing the 10^5 (10^4+10^4), to 10^4 ?
Long videos are needed, the problem is with ppls attention span, which is reduced to less than that of a goldfish due to scrolling reels.
Bro u said 35 k subs in beginning
the best! thank you
6:20 bro i moved on
how java guy can play with bitsets
❣❣❣❣
bhai koi aur color ki t shirt plz.
Much love :)
You are always my last option to understand a problem, not because you are bad at teaching, rather it means no it's time that I need my best weapon against that question.
i was trying to do dp with binary search, because we require next Element to be greater than the previous one,
i tried to do like this, jaise aapne ISS VIDEO mein bataya tha,
bhaiya aap bata sakte ho ki iss question mein ye solution work kyo nahi kar raha.
czcams.com/video/Uz2iNhjGGm0/video.htmlsi=Tg7B_P-OXmPaHGDT
(edit) :
int maxTotalReward(vector& nums) {
// Remove duplicates and sort the nums vector
sort(nums.begin(), nums.end());
nums.erase(unique(nums.begin(), nums.end()), nums.end());
int n = nums.size();
// Define the solve function with memoization
function solve = [&](int t) -> int {
if (t == 0) {
return 0;
}
auto p = lower_bound(nums.begin(), nums.end(), t);
if (p == nums.end()) {
--p;
}
if (*p == t) {
return t;
} else if (*p > t) {
if (p != nums.begin()) {
--p;
} else {
return 0;
}
}
int ans = 0;
while (p >= nums.begin() && *p * 2 - 1 > ans) {
ans = max(*p + solve(min(t - *p, *p - 1)), ans);
if (p == nums.begin()) {
break;
}
--p;
}
return ans;
};
return solve(nums.back() * 2 - 1);
}
i found this binary search solution but not able to understand it, can anyone please tell why the call is from (nums.back() * 2 - 1); and not (nums.back() - 1); simply
samjh nhi ata, question itna hard hai ya, ye banda hard bana ke samjhata hai ..😂
Mujhe pata hai tumhe kyu nahi samajh aa raha, aisa isliye ho raha kyuki shayad tum hi gadhe ho