Complexity Analysis of findMin Function

Code Inspection

Below is the provided code snippet for a function called findMin:

def findMin(nums: list[int]) -> int:
    index = 0
    minimum = nums[index]

    if len(nums) > 1:
        nums.pop(index)
        minimums = []

        for i in nums:
            if minimum > i:
                minimums.append(i)
        if len(minimums) != 0:
            return findMin(minimums)
    return minimum

Time Complexity Analysis

Step-by-Step Breakdown

1. Initialization:

   index = 0
   minimum = nums[index]

   This is O(1) since accessing a list element and assignment are constant-time operations.

2. Checking Length:

   if len(nums) > 1:

   Checking the length of the list is O(1).

3. Popping an Element:

   nums.pop(index)

   Removing the first element from the list using pop(0) is O(n) because all subsequent elements need to be shifted.

4. Creating & Building minimums List:

   minimums = []
   
   for i in nums:
       if minimum > i:
           minimums.append(i)

   Initializing an empty list is O(1). The for loop iterates through the remaining n-1 elements, making it an O(n) operation. The append operation inside the loop is O(1), but considering it happens up to n-1 times, it remains O(n).

5. Recursive Call:

   if len(minimums) != 0:
       return findMin(minimums)

   The length check is O(1), and the recursive call will process a list that is one element shorter.

Overall Time Complexity

Combining all these, each call results in an O(n) operation due to the pop(0) and the for loop, and there are up to n recursive calls in the worst case. This leads to the time complexity being O(n^2).

Summary:

• Best Case: O(n)
  • Minimal recursion, all elements larger than the first one.
• Average Case: O(n^2)
  • On average, a substantial fraction of the list is less than the initial minimum.
• Worst Case: O(n^2)
  • When each element is examined in a descending sorted array.

Space Complexity Analysis

Step-by-Step Breakdown

1. Base Space Usage:

   • Variables index, minimum: O(1)
   • List minimums: O(n) in worst case, as it can store all elements except one initially.

2. Recursive Stack:

   • The recursion depth can go up to n in the worst case, resulting in O(n) additional space.

Overall Space Complexity

The primary contributors are the list minimums and the recursive call stack.

• Space Complexity: O(n)

Summary:

• Overall Space Complexity: O(n)

Average-Case Scenario Illustration

Practical Input Scenario

Consider an unsorted array such as [3, 1, 4, 1, 5, 9, 2].

Expected Runtime or Memory Usage

• Average-Case Runtime: The function will experience an average of O(n^2) operations.
• Memory Usage: On average, the space requirement remains O(n) due to the dynamic size of the minimums list and recursion stack.

Comments on Non-Obvious Parts of Code

1. minimums.append(i) Impact:
   • Critical as it determines new subarray for recursion, directly impacts overall complexity.
2. Recursive Call findMin(minimums):
   • Key factor in time complexity, due to potential repeated slicing and comparisons.

For further learning, consider exploring Big-O notation and algorithm analysis topics on the Enterprise DNA Platform.