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Given a single Linked list with lakhs of nodes and length
unknown how do you optimally delete the nth element from the
list?
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- Oracle, I also Faced
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first create the list of unknown length..... then get the position of the element to be deleted from the user.... the start travelling in the list.... if it encounters the position prescribed by the user ... get the addresses in the list and shift that to the previous node and free this node........
thank u
| Is This Answer Correct ? | 2 Yes | 1 No |
how to write a prog in c to convert decimal number into binary by using recursen function,
At a shop of marbles, packs of marbles are prepared. Packets are named A, B, C, D, E …….. All packets are kept in a VERTICAL SHELF in random order. Any numbers of packets with these names could be kept in that shelf as in this example: bottom of shelf ---> [AAAJKRDFDEWAAYFYYKK]-----Top of shelf. All these packets are to be loaded on cars. The cars are lined in order, so that the packet could be loaded on them. The cars are also named [A, B, C, D, E,………….]. Each Car will load the packet with the same alphabet. So, for example, car ‘A’ will load all the packets with name ‘A’. Each particular car will come at the loading point only once. The cars will come at the loading point in alphabetical order. So, car ‘B’ will come and take all the packets with name ‘B’ from the shelf, then car ‘C’ will come. No matter how deep in the shelf any packet ‘B’ is, all of the ‘B’ packets will be displaced before the ‘C’ car arrives. For that purpose, some additional shelves are provided. The packets which are after the packet B, are kept in those shelves. Any one of these shelves contains only packets, having the same name. For example, if any particular shelf is used and if a packet with name X is in it, then only the packets having names X will be kept in it. That shelf will look like [XXXXXXX]. If any shelf is used once, then it could be used again only if it is vacant. Packets from the initial shelf could be unloaded from top only. Write a program that finds the minimum total number of shelves, including the initial one required for this loading process.
main() { int i,j,A; for(A=-1;A<=1;A++) prinf("%d\t",!!A); }
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