Software Developer

Step 1) Have a visited flag with each node.
Step 2) Traverse the linked list and keep marking visited nodes.
Step 3) If you see a visited node again then there is a loop. This solution works in O(n) but requires additional information with each node.

Step 1) Let the given element be x, create two variable i = 0, j = n-1 as index of row and column
Step 2) Run a loop until i = n
Step 3) Check if the current element is greater than x then decrease the count of j. Exclude the current column.
Step 4) Check if the current element is less than x then increase the count of i. Exclude the current row.
Step 5) If the element is equal, then print the position and end.

Step 1) Calculate the length of the Linked List. Let the length be len. 
Step 2) Print the (len – n + 1)th node from the beginning of the Linked List.

Step 1) There is N/2 squares or cycles in a matrix of side N. Process a square one at a time. Run a loop to traverse the matrix a cycle at a time, i.e loop from 0 to N/2 – 1, loop counter is i
Step 2) Consider elements in group of 4 in current square, rotate the 4 elements at a time. So the number of such groups in a cycle is N – 2*i.
Step 3) So run a loop in each cycle from x to N – x – 1, loop counter is y
Step 4) The elements in the current group is (x, y), (y, N-1-x), (N-1-x, N-1-y), (N-1-y, x), now rotate the these 4 elements, i.e (x, y) <- (y, N-1-x), (y, N-1-x)<- (N-1-x, N-1-y), (N-1-x, N-1-y)<- (N-1-y, x), (N-1-y, x)<- (x, y)
Step 5) Print the matrix.