Below is a clear, beginner-friendly explanation with complete Python implementations for Options A, B, C, and D, all of which:

• Preserve the original order

• Work with positive, negative, and zero values

• Do NOT use <code>set()</code> or <code>dict.fromkeys()</code>

• Manually handle duplicate detection

Problem Recap

Input:

[1, 2, 2, 3, 1]

Output:

[1, 2, 3]

We keep the first appearance only and ignore later duplicates.

✅ Option A: Loop + Extra List (Seen Values)

Idea

• Maintain a <code>seen</code> list to remember numbers already encountered

• Loop through the input list

• If a number is not in <code>seen</code>, add it to both <code>seen</code> and <code>result</code>

Why it works

• Order is preserved

• Easy to understand

• Time complexity: O(n²) (because <code>in</code> on a list is linear)

Code

def unique_elements_option_a(lst):
    seen = []
    result = []
    for num in lst:
        if num not in seen:
            seen.append(num)
            result.append(num)
    return result

✅ Option B: Nested Loop (No Extra Data Structure)

Idea

• For each element, check all previous elements

• If it appeared before, skip it

• Otherwise, add it to the result

Why it works

• Uses only loops

• No extra memory for tracking duplicates

• Time complexity: O(n²)

Code

def unique_elements_option_b(lst):
    result = []
    for i in range(len(lst)):
        duplicate = False
        for j in range(i):
            if lst[i] == lst[j]:
                duplicate = True
                break
        if not duplicate:
            result.append(lst[i])
    return result

✅ Option C: Using Index Comparison (<code>list.index()</code>)

Idea

• <code>lst.index(x)</code> gives the first occurrence of <code>x</code>

• If the current index equals the first index → keep it

• Otherwise → duplicate

Why it works

• Very readable

• Still avoids <code>set()</code> and <code>dict</code>

• Time complexity: O(n²)

Code

def unique_elements_option_c(lst):
    result = []
    for i in range(len(lst)):
        if lst.index(lst[i]) == i:
            result.append(lst[i])
    return result

✅ Option D: Build Result List & Check Against It

Idea

• Build the result list gradually

• Only add a number if it’s not already in the result

Why it works

• Simple and clean

• Preserves order naturally

• Time complexity: O(n²)

Code

def unique_elements_option_d(lst):
    result = []
    for num in lst:
        if num not in result:
            result.append(num)
    return result

🔍 Example Usage (Same for All Options)

data = [1, 2, 2, 3, 1]
print(unique_elements_option_a(data))
print(unique_elements_option_b(data))
print(unique_elements_option_c(data))
print(unique_elements_option_d(data))

Output:

[1, 2, 3]

🧠 Summary Table