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I think there is an edge case in 7 - 8 - 9 - 10 - 7 - 9. We can remove 7-8 and then 8-9 then we have 1 more component = 8. So we don't really need to remove 2 edges to get one more component. Instead I would find the bridges. If a + bridges >= k thats a return. Else keep track of the sizes of cycles like in this case 3, 3. Then individually check for those cyclic components
Not sure about Cassandra, I feel when we have a very complex search query structure like - filter on age, location, interests, gender, hobbies etc a search engine that supports text based search (basically quick indexing on props) like ES may be an option. But Cassandra looks like a horrible choice for a read-heavy system like Tinder :( Surprised to see Gaurav was okay with this lol
orderValidator is a function related to order, we can have it in orderManager/orderService only, it should not be considered as violation of SRP. or am i getting it wrong.
Suppose there are 2 seats available 2 users want to book 1 seat each at 1 sec gap lets say so when first user starts booking, u will lock all the seats?? NO right?? so i think we will need multiple enteries for seats rather than one +there are different kind of berths etc + quotas (senior citizen, ladies, gn, foreign, tq) so it is not that easy in real life🤯
This can also be one solution 1. Sort the array 2. Find min as a[0] 3. Subtract min-1 to all numbers 4. Now we have min as 1 and our max can he n. 5. Make another array as visited. 6. Ittrate in the array and for each value in the array mark visited[value] =1. 7. At last count number of zeros in visited array. Is it correct?? 😊
Problem very similar to travelling salesman problem and he is grandmaster it would be kpop for him 😂 but the way he came up with soln and just code it with no problems .. i mean thats incredible
for the first question i think two pointer approach would work pushing all index of thiefs in a vector and separate vector for police then try to give a thief who is as left as possible and if you cant then go for next atmost k difference of distance from the current police is that it works or not mam
