Insertion sort is a straightforward and efficient sorting algorithm for small datasets. It builds the final sorted array one element at a time. It is much like sorting playing cards in your hands: you take one card at a time and insert it into its correct position among the already sorted cards.

**Algorithm Overview:**

- **Start from the Second Element:** Begin with the second element, assuming the first element is already sorted.

- **Compare with Sorted Subarray:** Take the current element and compare it with elements in the sorted subarray (the part of the array before the current element).

- **Insert in Correct Position:** Shift all elements in the sorted subarray that are greater than the current element to one position ahead. Insert the current element into its correct position.

- **Repeat Until End:** Repeat this process for all elements in the array.

Let's sort the list `[5, 3, 8, 1, 2]` using insertion sort.

**Step-by-Step Visualization:**

**Initial List:** `[5, 3, 8, 1, 2]`

1. **Pass 1:**

- Current element: 3

- Compare 3 with 5, move 5 to the right: `[5, 5, 8, 1, 2]`

- Insert 3 in its correct position: `[3, 5, 8, 1, 2]`

2. **Pass 2:**

- Current element: 8

- 8 is already in the correct position: `[3, 5, 8, 1, 2]`

3. **Pass 3:**

- Current element: 1

- Compare 1 with 8, move 8 to the right: `[3, 5, 8, 8, 2]`

- Compare 1 with 5, move 5 to the right: `[3, 5, 5, 8, 2]`

- Compare 1 with 3, move 3 to the right: `[3, 3, 5, 8, 2]`

- Insert 1 in its correct position: `[1, 3, 5, 8, 2]`

4. **Pass 4:**

- Current element: 2

- Compare 2 with 8, move 8 to the right: `[1, 3, 5, 8, 8]`

- Compare 2 with 5, move 5 to the right: `[1, 3, 5, 5, 8]`

- Compare 2 with 3, move 3 to the right: `[1, 3, 3, 5, 8]`

- Insert 2 in its correct position: `[1, 2, 3, 5, 8]`

def insertion_sort(arr):

# Traverse from 1 to len(arr)

for i in range(1, len(arr)):

key = arr[i]

# Move elements of arr[0..i-1], that are greater than key,

# to one position ahead of their current position

j = i - 1

while j >= 0 and key < arr[j]:

arr[j + 1] = arr[j]

j -= 1

arr[j + 1] = key

arr = [5, 3, 8, 1, 2]

insertion_sort(arr)

print("Sorted array:", arr) # Output: [1, 2, 3, 5, 8]

- **Worst Case:** `𝑂(𝑛^2)` comparisons and swaps. This occurs when the array is in reverse order.

- **Best Case:** `𝑂(𝑛)` comparisons and `𝑂(1)` swaps. This happens when the array is already sorted.

- **Average Case:** `𝑂(𝑛^2)` comparisons and swaps. This is the expected number of comparisons and swaps over all possible input sequences.

Heap Sort is an efficient comparison-based sorting algorithm that uses a binary heap data structure. It divides its input into a sorted and an unsorted region and iteratively shrinks the unsorted region by extracting the largest (or smallest) element and moving it to the sorted region.

**Algorithm Overview:**

- **Build a Max Heap:** Convert the array into a max heap, a complete binary tree where the value of each node is greater than or equal to the values of its children.

- **Heapify:** Ensure that the subtree rooted at each node satisfies the max heap property. This process is called heapify.

- **Extract Maximum:** Swap the root (the maximum element) with the last element of the heap and reduce the heap size by one. Restore the max heap property by heapifying the root.

- **Repeat:** Continue extracting the maximum element and heapifying until the entire array is sorted.

Let's sort the list `[5, 3, 8, 1, 2]` using heap sort.

1. **Build Max Heap:**

- Initial array: `[5, 3, 8, 1, 2]`

- Start heapifying from the last non-leaf node.

- Heapify at index 1: `[5, 3, 8, 1, 2]` (no change, children are already less than the parent)

- Heapify at index 0: `[8, 3, 5, 1, 2]` (swap 5 and 8 to make 8 the root)

2. **Heapify Process:**

- Heapify at index 0: `[8, 3, 5, 1, 2]` (no change needed, already a max heap)

3. **Extract Maximum:**

- Swap root with the last element: `[2, 3, 5, 1, 8]`

- Heapify at index 0: `[5, 3, 2, 1, 8]` (swap 2 and 5)

4. **Repeat Extraction:**

- Swap root with the second last element: `[1, 3, 2, 5, 8]`

- Heapify at index 0: `[3, 1, 2, 5, 8]` (swap 1 and 3)

- Swap root with the third last element: `[2, 1, 3, 5, 8]`

- Heapify at index 0: `[2, 1, 3, 5, 8]` (no change needed)

- Swap root with the fourth last element: `[1, 2, 3, 5, 8]`

After all extractions, the array is sorted: `[1, 2, 3, 5, 8]`.

def heapify(arr, n, i):

largest = i # Initialize largest as root

left = 2 * i + 1 # left child index

right = 2 * i + 2 # right child index

# See if left child of root exists and is greater than root

if left < n and arr[largest] < arr[left]:

largest = left

# See if right child of root exists and is greater than root

if right < n and arr[largest] < arr[right]:

largest = right

# Change root, if needed

if largest != i:

arr[i], arr[largest] = arr[largest], arr[i] # swap

# Heapify the root.

heapify(arr, n, largest)

def heap_sort(arr):

n = len(arr)

# Build a maxheap.

for i in range(n // 2 - 1, -1, -1):

heapify(arr, n, i)

# One by one extract elements

for i in range(n - 1, 0, -1):

arr[i], arr[0] = arr[0], arr[i] # swap

heapify(arr, i, 0)

arr = [5, 3, 8, 1, 2]

heap_sort(arr)

print("Sorted array:", arr) # Output: [1, 2, 3, 5, 8]

- **Worst Case:** `𝑂(𝑛log𝑛)`. Building the heap takes `𝑂(𝑛)` time, and each of the 𝑛 element extractions takes `𝑂(log𝑛)` time.

- **Best Case:** `𝑂(𝑛log𝑛)`. Even if the array is already sorted, heap sort will still build the heap and perform the extractions.

- **Average Case:** `𝑂(𝑛log𝑛)`. Similar to the worst-case, the overall complexity remains `𝑂(𝑛log𝑛)` because each insertion and deletion in a heap takes `𝑂(log𝑛)` time, and these operations are performed 𝑛 times.