System Engineer

Task 1: Read one coin at a time in the loop. Take its value. Let’s call this a[j].
Task 2: In an inner loop, go through each coin in the list and count how many times is V occurring. For this we first need to initialize count to zero. Whenever a[j] == a[i] is true, we need to increment the counter.
Task 3: Once the inner loop is completed, the value of count will tell us how many times has a[i] occurred in the array. 
Task 4: We need to check if the count is Odd. We are told that only 1 denomination will occur an odd number of times. If we find it, we can print it and exit the program.
When we divide an Odd number with 2, we get 1 as remainder. This can be obtained using the % operator.

Mergesort is a popular divide-and-conquer sorting algorithm known for its stability and efficiency. It works by recursively dividing the unsorted list into smaller sublists until each sublist contains only one element. Then, it merges these sublists in a sorted manner to produce the final sorted list. Here's a step-by-step approach to solving Mergesort:

Base Case:

• If the input list has zero or one element, it is already considered sorted. Return the list as it is.

Divide:

• Divide the input list into two equal-sized sublists. You can do this by finding the middle index of the list.

Recursively Sort:

• Recursively apply the Mergesort algorithm to both sublists generated in the previous step. This will continue until you reach the base case, where the sublists contain only one element.

Merge:

• Merge the two sorted sublists from the previous step into a single sorted list. To do this, compare elements from both sublists and merge them in ascending order into a new list. Continue this process until you have merged all elements from both sublists.

Return the Result:

• The merged list from the previous step is now a sorted list containing all the elements from the original input list. Return this sorted list as the final result.