similar question on leetcode -> 1552. Magnetic Force Between Two Balls 1870. Minimum Speed to Arrive on Time 875. Koko Eating Bananas 1011. Capacity To Ship Packages Within D Days 1283. Find the Smallest Divisor Given a Threshold 1482. Minimum Number of Days to Make m Bouquets 2064. Minimized Maximum of Products Distributed to Any Store
Thank you sir for this video. This question was not clear for me after watching a video on this question on TH-cam. But your explanation is extraordinary ❤️🔥. It's crystal clear for me now. Keep making more these top notch quality content. ❤️❤️🔥.
Striver is the dsa playlist the solns for the sde sheet that u prepared.For example does the linked list playlist consists of the solutions to the linked list problems in the sde sheet.Please answer because if they are the solns i ll watch them after trying the sde sheet.
we can also take low as the minimum gap between the consecutive element alfter sorting. For example if the array is 10 50 100 200. then we can take low as 40. because that can be the minimum ans if the number of cows is 4. This will reduce the time complexity for binary search. Please correct me if i am wrong.
yeah, ig coz in book allocation problem the search space was from max_element ---- sum of all elements altho here we are not finding the sum as the problem statement is diff we need the min distance that is low till the end
More Good Binary Search Problems (Leetcode). Good luck and have fun! 1539. Kth Missing Positive Number 1482. Minimum Number of Days to Make m Bouquets 1283. Find the Smallest Divisor Given a Threshold 1231. Divide Chocolate 1011. Capacity To Ship Packages In N Days 875. Koko Eating Bananas 774. Minimize Max Distance to Gas Station 410. Split Array Largest Sum
Everything in this video is clear and excellent. However I couldn't figure out the below. Can Anyone explains me to clear it out. Why we need to place our first cow in first stall. Placing the first cow in different(next) stalls leads to different answer right? Also why are we assuming min distance of cow as firstly placed two cows, i.e., stall[0] and next upcoming stall[i] that covers min distance, probably min distance will be on next upcoming stalls as well right? Hoping to get clear my doubts. Thanks.
Thanks for those 2 Gems... those are really something that I am searching for . Both in book allocation and Aggressive cows question's, you give a quality explanation that Isolved some similar question like that by my own. Thank you.
Hi! Please upload a video on Time and Space Complexity if possible . I just finished watching your C++ STL video today. You're doing a great job👏. I really like the way you explain things🥺🔥 Thanks a ton 🙏❤️
Can someone explain why we are always assuming cow1 to be at index 0. At 19:30 mins in video co-ord = a[0] ? Can there not be any case when 1st cow is placed at any other index other than zero ?
We need the best solution, thus the best solution is to place first cow in the first stall, and then from onwards we will get the best solution automatically.
Haye your excitement in 10:52, excites me I get more Interested in Q after seeing your interest and energy, which matters me the most for my love of coding. You're true gems in our community I seriously admire your nature and thought process.❤️
Thanx Striver Bhaiya.came here from your Binary Search master class and I easily solved all the below listed questions .Thanx for these amazing concepts
why is lowest possible distance 1 ? lowest possible should be when 2 cows are placed sided by side in the first 2 stalls with minimum values, but no body said the distance between first 2 stalls will be guaranteed 1, what if the array is 1 3 5 7 9, will we ever have a min distance 1 b/w 2 cows in such a case ? I think it should be arr[1] - arr[0] after sorting the array. Can somebody explain ?
Exactly ! that's where I'm stuck . Like there could be other combination from the list which might get possible for given mid . Please can anyone explain this
Because we want to know whether we can place the cows at a min distance of 'dist'. So, if by placing the cow at the first co-ordinate, we cannot get a configuration where the dist between consecutive cows is >= dist, then by placing the first cow at any other position as well we wont get any configuration
i invest my time to understand 7 hours i have made many times dry and run i try to solve it using sliding window and brute fornce and binary search but i couldn't now i understood complitely thank u bro thank u so mcuh ........
Ek dum mast sequential way mein aa rha hai video pura thought process bana diye hai pura relate kar paye pichle video se pehle toh q dekh ke pata hi nhi chalta tha ki binary search ka q hai 😊
@@naveensingh2658 why, lowest possible should be when 2 cows are placed sided by side in the first 2 stalls with minimum values, but no body said the distance between first 2 stalls will be guaranteed 1, what if the array is 1 3 5 7 9, will we ever have a min distance b/w 2 cows in such a case ? I think it should be arr[1] - arr[0] after sorting the array.
@@cenacr007 No actually there could be a case for ex. - 3 6 9 10 11 13 where we need to place 6 cows then min possible would be 1 in that case so in my opinion it should be actually min(arr[i + 1], arr[i]) for i = 0 till size - 2 index (in this case till 4th index) but it doesn't matter even if we take 1 as the minimum because the final answer will only give the maximum possible minimum distance.
Great explanation. I understand that we have to be greedy and place the first cow always at the first position, but if after doing that we do not arrive at the solution, can no other solution exist if place the first cow at 2nd/3rd position and then try to place rest of the cows?
what if instead of 9 there is 12. now if we try for ans 4 c1 = 1 c2 = 8 c3 = 12 c1-c2 = 7 c2-c3 = 4 we can get ans for 4 , by evaluating c2-c3. if we only check c1-c2, then i guess we'll miss this testcase. (instead of 9, 12)
After lot of struggle through many videos finally here I understood this concept 🥰. Kisko credit diya jaye meri mehanat ko ya striver bhaiya❤ ko. yrr sach me book allocation problem ke bad 2 dino se samjhne ki koshish kar rha tha🤒😅
Hi, I know you are busy, recently i have gone through you segment tree with lazy propagation lecture, the concept u provide is amazing. Can you please make a video on persistent segment tree when you are free?
I have one doubt, you said if it's an unsorted array then sort it . Then after sorting ,the distance between the coordinates will change and perhaps answer will also change
What If we wanted to find smallest minimum distance in this problem? I think instead of going only to mid + 1 to high in success scenario at mid we must explore low to mid - 1..... M I right plz answer this?
@@saurabhanand6590 you cannot directly apply a single lower bound.. you have to apply for every cow bro.. so takes more time rather.. For C cows it becomes C*log n
Bhaiya it would be great if you can also tell the companies in which the questions are asked. Many of us don't even have a clue of frequency and company in which questions are mostly asked. It helps a lot
If the array is [16,17,18,0] and we have to place two cows then how will give the right answer as we will always start with a[0] i.e 16 and can never get 18,0 as answer which is the largest smallest value possible
Yeah I got it afterwards, we have to just do the inorder traversal and then keep on counting the nodes which we have visited till now. Whenever we visit a node we increment the count and then we check if that count is equal to k. If it is, then we return that element otherwise we keep on doing the traversal.
Please watch our new video on the same topic: th-cam.com/video/R_Mfw4ew-Vo/w-d-xo.html
Problem in leetcode : Magnetic force between 2 balls. Leetcode...
Aggressive Striver :)
😂
similar question on leetcode ->
1552. Magnetic Force Between Two Balls
1870. Minimum Speed to Arrive on Time
875. Koko Eating Bananas
1011. Capacity To Ship Packages Within D Days
1283. Find the Smallest Divisor Given a Threshold
1482. Minimum Number of Days to Make m Bouquets
2064. Minimized Maximum of Products Distributed to Any Store
Thank you man. Really appreciate it
@@marvel438 welcome
1231 Divide Chocolate
774 Minimize Max Distance to Gas Station
410 Split Array Largest Sum
these too :D
nice, thanks!
Thankss!
for all those who are confuse , at 3:21 the c2 should be drawn under the stall 8 column not under stall 4 , although explanation was good
Your explanaining capabilities are absolutely beautiful. People really be coding it out without thinking the approach which you did beautifully.
Checkout out playlist for all DSA in the description.
Follow me at Instagram for updates: Striver_79
Thank you sir for this video. This question was not clear for me after watching a video on this question on TH-cam.
But your explanation is extraordinary ❤️🔥. It's crystal clear for me now.
Keep making more these top notch quality content. ❤️❤️🔥.
Hindi mein bbhi pdahye sir
high=arr[n-1]-arr[0]
Striver is the dsa playlist the solns for the sde sheet that u prepared.For example does the linked list playlist consists of the solutions to the linked list problems in the sde sheet.Please answer because if they are the solns i ll watch them after trying the sde sheet.
Aggressive cows , explained in a calm manner, striver bhaiya op🔥
He was also agressive in making solution understood :)
we can also take low as the minimum gap between the consecutive element alfter sorting. For example if the array is
10 50 100 200.
then we can take low as 40. because that can be the minimum ans if the number of cows is 4. This will reduce the time complexity for binary search. Please correct me if i am wrong.
yeah, ig coz in book allocation problem the search space was from max_element ---- sum of all elements
altho here we are not finding the sum as the problem statement is diff we need the min distance that is low till the end
Thank you bhaiya for this amazing explanation. But, the "NOOO! WHY?" at 10:40 is epic🤣
More Good Binary Search Problems (Leetcode).
Good luck and have fun!
1539. Kth Missing Positive Number
1482. Minimum Number of Days to Make m Bouquets
1283. Find the Smallest Divisor Given a Threshold
1231. Divide Chocolate
1011. Capacity To Ship Packages In N Days
875. Koko Eating Bananas
774. Minimize Max Distance to Gas Station
410. Split Array Largest Sum
thanks a lot for sharing
Everything in this video is clear and excellent.
However I couldn't figure out the below. Can Anyone explains me to clear it out.
Why we need to place our first cow in first stall. Placing the first cow in different(next) stalls leads to different answer right?
Also why are we assuming min distance of cow as firstly placed two cows, i.e., stall[0] and next upcoming stall[i] that covers min distance, probably min distance will be on next upcoming stalls as well right?
Hoping to get clear my doubts. Thanks.
Incredible Explanation, Thanks for the Tutorial!
Your explanation and the topic clearance has grown exponentially with whiteboard, you are gonna progress a lot for sure , no looking back.
Thanks for those 2 Gems... those are really something that I am searching for .
Both in book allocation and Aggressive cows question's, you give a quality explanation that Isolved some similar question like that by my own. Thank you.
Hi! Please upload a video on Time and Space Complexity if possible . I just finished watching your C++ STL video today. You're doing a great job👏. I really like the way you explain things🥺🔥 Thanks a ton 🙏❤️
Can someone explain why we are always assuming cow1 to be at index 0. At 19:30 mins in video co-ord = a[0] ? Can there not be any case when 1st cow is placed at any other index other than zero ?
We need the best solution, thus the best solution is to place first cow in the first stall, and then from onwards we will get the best solution automatically.
Bro i knew this prblm and still came here see the way u explain, and just after starting hit the like 💯💯💯. Great wrk bro
Haye your excitement in 10:52, excites me I get more Interested in Q after seeing your interest and energy, which matters me the most for my love of coding. You're true gems in our community I seriously admire your nature and thought process.❤️
Bro the way u teach, the energy, makes learning so fun and interesting....tysm Brother!
Your hard work and dedication can be clearly seen from ur eyes🙏🙏🙏
This is honestly one of the best explanation for all the questions I came across ever on youtube. Thanks sir for such awesome content.
Thanx Striver Bhaiya.came here from your Binary Search master class and I easily solved all the below listed questions .Thanx for these amazing concepts
It was pretty tough to figure out the search space right after going through the question, but your intuition was really good.
why is lowest possible distance 1 ? lowest possible should be when 2 cows are placed sided by side in the first 2 stalls with minimum values, but no body said the distance between first 2 stalls will be guaranteed 1, what if the array is 1 3 5 7 9, will we ever have a min distance 1 b/w 2 cows in such a case ? I think it should be arr[1] - arr[0] after sorting the array. Can somebody explain ?
same doubt bhai
yeah the search space will be more specifically low will be after sorting arr[1] - arr[0]
18:58
How to place cows;
we're finding " do ke bich ki duri jitna ho sake badha do., but within given stalls".
Bhaiya classes k din yaad a gye..... Loved it...... Nd all thanks to u for making such a tough prblm bread and butter.....
Kindly include painters partition too please.
Its same as allocation of minimum pages, exactly same problem..
I don't understand, why the first cow always has to be plased at the first given co-ordinate?
Exactly ! that's where I'm stuck . Like there could be other combination from the list which might get possible for given mid . Please can anyone explain this
Because we want to know whether we can place the cows at a min distance of 'dist'.
So, if by placing the cow at the first co-ordinate, we cannot get a configuration where the dist between consecutive cows is >= dist, then by placing the first cow at any other position as well we wont get any configuration
3:20 konse nashe h? How the hell difference of 2 and 4 is 6😢 and difference of 4 and 9 is 1
Slider corrected it..
@@takeUforward btw you explain things very well.. abhi tak 2 videos dekhi h lekin concepts bahut badhiya or clarity ke saath smjh aajate h..
i invest my time to understand 7 hours i have made many times dry and run i try to solve it using sliding window and brute fornce and binary search but i couldn't now i understood complitely thank u bro thank u so mcuh ........
Ek dum mast sequential way mein aa rha hai video pura thought process bana diye hai pura relate kar paye pichle video se pehle toh q dekh ke pata hi nhi chalta tha ki binary search ka q hai 😊
Why does the first cow have to be at the left extreme?
Is this question a part of the SDE sheet, because I can't find it in any section.
Anyway, Kudos to your efforts!
Binary search, please use the google doc version, and don't download it!! As I keep updating it..
@@takeUforward kuch ladke humare pehchan ke hardcopy nikaal ke highlighter chalate hai sde sheet pe unka kya kiya jaye😂
@@abhishekkumarsingh8603 hahaahahahaahahah
intuition behind each step nicely explained.
there is a drastic improvement from video-1 of whiteboard. you are improving by each video.
I did not understand, taking arr[0] for the first cow.(Greedy Approach). Please explain
Please, can anyone explain, why to take low=1, instead of low=xi[1]-xi[0] in binary search approach.
ThankYou in advance
1 will be the lowest possible minimum distance.
@@naveensingh2658 why, lowest possible should be when 2 cows are placed sided by side in the first 2 stalls with minimum values, but no body said the distance between first 2 stalls will be guaranteed 1, what if the array is 1 3 5 7 9, will we ever have a min distance b/w 2 cows in such a case ? I think it should be arr[1] - arr[0] after sorting the array.
@@cenacr007 No actually there could be a case for ex. - 3 6 9 10 11 13 where we need to place 6 cows then min possible would be 1 in that case so in my opinion it should be actually min(arr[i + 1], arr[i]) for i = 0 till size - 2 index (in this case till 4th index) but it doesn't matter even if we take 1 as the minimum because the final answer will only give the maximum possible minimum distance.
More accurately i say low is the minimum value among all the consecutive differences of elements.
Great explanation. I understand that we have to be greedy and place the first cow always at the first position, but if after doing that we do not arrive at the solution, can no other solution exist if place the first cow at 2nd/3rd position and then try to place rest of the cows?
I am a little confused abt this part
how do we set for a particular problem high = mid-1 or high = mid in some other problem and low
I think -
When you are using high = mid -1 and low = mid +1, then use low
Sir why are we always picking the first position as optimal in the canPlaceCows function?
at 14:02 why c1 and c2 is kept at 1 and 4 , it should be at 2 and 4
what if instead of 9 there is 12.
now if we try for ans 4
c1 = 1
c2 = 8
c3 = 12
c1-c2 = 7
c2-c3 = 4
we can get ans for 4 , by evaluating c2-c3.
if we only check c1-c2, then i guess we'll miss this testcase. (instead of 9, 12)
After lot of struggle through many videos finally here I understood this concept 🥰.
Kisko credit diya jaye meri mehanat ko ya striver bhaiya❤ ko. yrr sach me book allocation problem ke bad 2 dino se samjhne ki koshish kar rha tha🤒😅
Hi, I know you are busy, recently i have gone through you segment tree with lazy propagation lecture, the concept u provide is amazing. Can you please make a video on persistent segment tree when you are free?
Your excellent art of explaining these questions has inspired me to not lose hope and improve my problem solving skills further
how can you place three cows 10:52 mast bhaiya
The lesson learnt here is that : "Binary Search can also be used to find max value of a monotonic function" .....
I have one doubt, you said if it's an unsorted array then sort it . Then after sorting ,the distance between the coordinates will change and perhaps answer will also change
Nobody gives a damn about your Doubts!
Thanks to your previous videos, and intelligent ordering of questions in the SDE sheet, I did this on my own. Thanks bhai :)
Here after binary search master class, was able to do aggressive cows on my own.
Understood ❤❤ Great Explanation ..keep this coming
"Definitely not!"😅😂 Reminds me of Dhoni saying this :p
What If we wanted to find smallest minimum distance in this problem? I think instead of going only to mid + 1 to high in success scenario at mid we must explore low to mid - 1..... M I right plz answer this?
You are superb buddy ..so hard working where do you get this inspiration from ? Just finished looking the last video and new one is ready 😊
can we use lower bound in can place cows function??
This way the time complexity other than sorting will be O((logn)^2)
Yes you can, it will become a bit complex though. Also the worst case of that too is N. imagine cows=n and stalls being 1,2,3...
@@takeUforward Thanks for the reply... I actually implemented the approach on leetcode and its runtime is higher than the linear ones.
@@takeUforward The Worst case time for lower bound will be O(log n) as it uses binary search
@@saurabhanand6590 you cannot directly apply a single lower bound.. you have to apply for every cow bro.. so takes more time rather..
For C cows it becomes C*log n
@@takeUforward Yes got your point bro... Thanks again
Understood! Super awesome explanation as always, thank you very much!!
Bhaiya thanks for this video and Tree series kab tab ayegi please bta do placements aane vali hai hamari
Thank You So Much Striver Brother............🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
Another best one from the best 👍💯
Bhaiya it would be great if you can also tell the companies in which the questions are asked. Many of us don't even have a clue of frequency and company in which questions are mostly asked. It helps a lot
Sde sheet for that.
One of the best explanation thanks Striver ! ❤❤
thank you striver, i really appreciate that man!!
Gone through all other video on this question. But this one is amazing
If the array is [16,17,18,0] and we have to place two cows then how will give the right answer as we will always start with a[0] i.e 16 and can never get 18,0 as answer which is the largest smallest value possible
sorting is important, that's why i said...
@@takeUforward yeah...
Missed the way..
Thanks as always..
today's codestudio's moderate difficulty problem, and i understood the approach but somewhere I am struggling to code
For this question array must be sorted?
why we are placing 1st cow each time at 0th index. can someone help?
great explaination.Thankyou for providing this
why are wee sorting the stalls array??
Vry nice explanation got every bit and learn something new. tysm
to place cow at 1st position we need to sort array otherwise we wont get correct answer
amazing explanation mann!!!!!!!!!!!!!!!!! thank you so mucchhhh
Bhaiya ismai sort karna zaruri kyu h array ko ?? unsorted array par binary search on answer nhi nikal sakte??
The energy 👀⚡⚡
can we also start placing cows from backward like from largest to lowest
Thank you bhaiya. Great explanation
After getting initial intuition able to solve by myself,thanks striver
bhai mind blowing your efforts in this video make my dsa journey easy
love you so much bro.
keep making this type of video in **WHITE BOARD
why is the low 1 and not min. of the diff?
Possible combinations are 5c3 i.e 10
great explanation sir
bhaiya is question me array jo milega wo sorted format me milega?
This is a very good explanation
Amazing explanation striver❤️👍🏻
Like the thumbnail simple 👌 just need to show front face.😄🏋🔥🔥🔥
Another masterpiece... ❤from CGEC bhaiya
Bhaiya kaya explain kiya ho !!! Lit !! Love this explanation :-)
similar to arrange books -> which are all based on painters partition alogo 😁👍
Beauty bhai! Jabardast explanation!
Thank you sir, great explanation
very lucidly explained!!
Very Nicely Explained :)
Your explanation is amazing sir
Poori B.Tech nikal gayi par ye question ek baar me kabhi solve nahi hua 😂. Ab MS karte hue dekh raha hun. Thanks bro😀
make a video on finding the kth smallest element in a binary search tree in O(N) time and O(1) space.
U just have to search that's it ... It remains O(n)
Yeah I got it afterwards, we have to just do the inorder traversal and then keep on counting the nodes which we have visited till now. Whenever we visit a node we increment the count and then we check if that count is equal to k. If it is, then we return that element otherwise we keep on doing the traversal.
Great explanation!
Why low is 1 ? can someone help me ?
Sir when will stack and queue solutions will come?
such clean explanation ❤