SDE - 1

We can easily solve this problem by following a Greedy approach. The idea is simple – consider each task decreasing order of their profits and schedule it in the latest possible free slot that meets its deadline. If no such slot is there, don’t schedule the task.

It can be solved using dynamic programming.
We will solve this problem in a bottom-up manner. In the bottom-up approach, we solve smaller subproblems first, then solve larger subproblems from them. The following bottom-up approach computes T[i], which stores the maximum profit achieved from the rod of length i for each 1 <= i <= n. It uses the value of smaller values I already computed.

The following steps are required for this Algorithm:
1. Create a dummy node and point it to the head of input i.e. dummy->next = head.
2. Calculate the length of the linked list which takes O(N) time, where N is the length of the linked list.
3. Initialize three-pointers prev, curr, next to reverse k elements for every group.
4. Iterate over the linked lists till next!=NULL.
5. Points curr to the prev->next and next to the curr next.
6. Then, Using the inner for loop reverse the particular group using these four steps:
curr->next = next->next
next->next = prev->next
prev->next = next
next = curr->next
7. This for loop runs for k-1 times for all groups except the last remaining element, for the last remaining element it runs 
for the remaining length of the linked list – 1.
8. Decrement count after for loop by k count -= k, to determine the length of the remaining linked list.

As the name suggests when the cache memory is full, LRU picks the data that is least recently used and removes it to make space for the new data. The priority of the data in the cache changes according to the need of that data i.e. if some data is fetched or updated recently then the priority of that data would be changed and assigned to the highest priority, and the priority of the data decreases if it remains unused operations after operations.
LRUCache (Capacity c): Initialize LRU cache with positive size capacity c.
get (key): Returns the value of Key ‘k’ if it is present in the cache otherwise it returns -1. Also updates the priority of data in the LRU cache.
put (key, value): Update the value of the key if that key exists, Otherwise, add a key-value pair to the cache. If the number of keys exceeds the capacity of the LRU cache then dismiss the least recently used key.