Is the dataset sorted?

This is not required for the given problem, but a very simple algorithm to implement.

# A function to determine if an array of integer is sorted

def isSorted(array : list[int]) -> bool:
    if len(array) <= 1:
        return True
    else:
        for i in range(1, len(array)):
            if array[i-1] > array[i]:
                return False
        # end of for
        return True
# end of isSorted()

list1 = [1,2,3]
list2 = [3,1,4,1,5,9]
print(f"Is {list1} sorted? {isSorted(list1)}")
print(f"Is {list2} sorted? {isSorted(list2)}")

Function Definition: isSorted

The function isSorted takes a list of integers as an input and returns a boolean indicating whether the list is sorted in non-decreasing order.

def isSorted(array: list[int]) -> bool:

• array: list[int]: This is a type annotation indicating that the parameter array is expected to be a list of integers.

• -> bool: This indicates that the function will return a boolean value (True or False).

1. Base Case:

if len(array) <= 1:
    return True

• If the length of the array is 0 or 1, the function returns True.

• A list with 0 or 1 element is trivially sorted because there are no pairs of elements to compare.

1. Loop Through the Array:

for i in range(1, len(array)):
    if array[i - 1] > array[i]:
        return False

• The function iterates through the array starting from the second element (i = 1) to the last element (i = len(array) - 1).

• Comparison:

  • For each index i, it compares the element at array[i-1] with the element at array[i].

  • If array[i-1] is greater than array[i], the function returns False immediately, indicating that the list is not sorted in non-decreasing order.

1. End of the For Loop:

return True

• If the loop completes without finding any elements that are out of order, the function returns True, indicating that the list is sorted.