That prevents the last sublist size from exceeding sqrtn, so it has a slightly better space utilization. You may want to round up using sqrtn = ceil(sqrt(n)). This program uses \$O(\sqrt n)\$ space because it uses one array of that size and one temporary list of that size. Static List *createReversedList(const List *head, int size) When writing the complexity, there are some rules to follow. List *tempList = createReversedList(subList, subListSize) Stack Overflow for Teams is moving to its own domain When the migration is complete, you will access your Teams at, and they will no longer appear in the left sidebar on. The last subList size may be different. Make a reversed copy of each subList and print it. What is the time complexity of the program to reverse stack when linked list is used for its implementation A. Remember the head of each sqrtn sized sublist Here is what the program would look like: void display_reverse(const List *list) To do that, you will need to use a \$\sqrt n\$ sized array to keep track of these sublists. If the list is \$n\$ items, you can make \$\sqrt n\$ sized sublist copies, and reverse/print them one at a time. A better way to create stacks and queues is using a. If you can't modify the list, you can do it by making copies of part of the list. When adding or removing, you may need to re-index the entire array, which can lead to O(N) time complexity. This whole function will take \$O(n)\$ time and \$O(1)\$ space. You will work with the best techies and professionals Our stack includes Python. Size of stack is: 5 Top element of stack is: 3 Performance Let n be the number of elements in the stack. All you need to do after that is print the list in order: void display_reverse(List *list) Responsibilities: You will be a part of the Axonius R
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