Algorithm efficiency basics (fast vs slow) in Intro to Computing - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the sum of numbers from 1 to n efficiently in O(1) time.

Intro to Computing

def sum_to_n(n):
    return n * (n [1] 1) // 2

print(sum_to_n(5))

Drag options to blanks, or click blank then click option'

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition.

Confusing the operators in the formula.

The formula n * (n + 1) // 2 computes the triangular number sum in constant O(1) time, faster than an O(n) loop for large n.

2fill in blank

medium

Complete the O(n) code to find the largest number in the list (linear scan).

Intro to Computing

numbers = [3, 7, 2, 9, 5]
max_num = numbers[0]
for num in numbers:
    if num [1] max_num:
        max_num = num
print(max_num)

Drag options to blanks, or click blank then click option'

A>

B<

C==

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' finds minimum instead.

Using '==' doesn't update properly.

A single loop scans the list once (O(n) time), efficient for finding the maximum without sorting O(n log n).

3fill in blank

hard

Complete the slow O(n^2) nested loop code to check for duplicates.

Intro to Computing

def has_duplicate(nums):
    for i in range(len(nums)):
        for j in range([1], len(nums)):
            if nums[i] == nums[j]:
                return True
    return False

print(has_duplicate([1,2,2,3]))

Drag options to blanks, or click blank then click option'

Ai

Bi+1

C0

Dlen(nums)

Attempts:

3 left

💡 Hint

Common Mistakes

Starting at 0 causes redundant checks.

Using i includes self-comparison.

Nested loops compare each pair (O(n^2) time), slow for large lists. Starting j at i+1 avoids self-comparison.

4fill in blank

hard

Complete the fast O(n) code using a set to check for duplicates.

Intro to Computing

def has_duplicate_fast(nums):
    seen = set()
    for num in nums:
        if num [1] seen:
            return True
        seen.[2](num)
    return False

print(has_duplicate_fast([1,2,2,3]))

Drag options to blanks, or click blank then click option'

Ain

Badd

C>

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong operator or method.

Forgetting to add after check.

Sets provide O(1) average lookup/add, making the whole algorithm O(n), much faster than O(n^2) nested loops.

5fill in blank

hard

Fill blanks for O(n) dict comprehension to map words to lengths (efficient single pass).

Intro to Computing

words = ['apple', 'bat', 'carrot', 'dog']
lengths = {word: [1] for word in words if len(word) [2] 3}
print(lengths)

Drag options to blanks, or click blank then click option'

Alen(word)

B>

C<=

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Wrong filter includes short words.

Using word as value instead of length.

Dict comprehension processes each word once O(n), faster in practice than equivalent loop with if checks.

Practice

1.

What does algorithm efficiency mainly measure?

easy

A. How fast or slow an algorithm solves a problem

B. The color of the computer screen

C. The size of the computer's hard drive

D. The number of users on a website

Solution

1. Step 1: Understand the meaning of algorithm efficiency

   Algorithm efficiency tells us how quickly or slowly an algorithm completes its task.

2. Step 2: Compare options to the definition

   Only How fast or slow an algorithm solves a problem matches the concept of speed or slowness of solving a problem.

3. Final Answer:

   How fast or slow an algorithm solves a problem -> Option A

4. Quick Check:

   Algorithm efficiency = speed of solving [OK]

Hint: Algorithm efficiency = speed of solving problems [OK]

Common Mistakes:

• Confusing efficiency with hardware specs
• Thinking efficiency is about user count
• Mixing efficiency with unrelated computer parts

2.

Which of these is a sign of a faster algorithm?

for i in range(n):
    print(i)

easy

A. The algorithm jumps directly to the middle item

B. The algorithm checks every item one by one

C. The algorithm repeats the same step many times

D. The algorithm uses more memory than needed

Solution

1. Step 1: Analyze the given code

   The code loops through all items from 0 to n-1, checking each one.

2. Step 2: Compare with options describing speed

   Jumping directly to the middle item is faster than checking all items one by one.

3. Final Answer:

   The algorithm jumps directly to the middle item -> Option A

4. Quick Check:

   Jumping steps = faster algorithm [OK]

Hint: Faster algorithms skip steps, not check all [OK]

Common Mistakes:

• Thinking looping over all items is fast
• Confusing memory use with speed
• Ignoring the benefit of skipping steps

3.

What is the output speed difference between these two algorithms when n is very large?

Algorithm 1: Check every item one by one
Algorithm 2: Jump to the middle, then half repeatedly

medium

A. Algorithm 1 is faster because it checks all items

B. Algorithm 2 is faster because it reduces steps quickly

C. Both algorithms take the same time

D. Algorithm 1 uses less memory so it is faster

Solution

1. Step 1: Understand the two algorithms

   Algorithm 1 checks all items one by one (slow for large n). Algorithm 2 jumps to the middle and halves the search repeatedly (fast for large n).

2. Step 2: Compare efficiency for large n

   Algorithm 2 reduces the number of steps quickly, making it faster than Algorithm 1.

3. Final Answer:

   Algorithm 2 is faster because it reduces steps quickly -> Option B

4. Quick Check:

   Halving steps = faster algorithm [OK]

Hint: Halving steps beats checking all [OK]

Common Mistakes:

• Assuming checking all is faster
• Ignoring step reduction benefits
• Confusing memory use with speed

4.

Find the error in this slow algorithm and suggest a faster approach:

def find_item(lst, target):
    for item in lst:
        if item == target:
            return True
    return False

medium

A. The algorithm has a syntax error in the loop

B. The algorithm uses too much memory; reduce list size

C. The algorithm returns the wrong value

D. The algorithm checks all items; use binary search on sorted list instead

Solution

1. Step 1: Identify the algorithm's behavior

   The function checks each item one by one until it finds the target or ends.

2. Step 2: Suggest a faster method

   Using binary search on a sorted list jumps to the middle and halves the search, making it faster.

3. Final Answer:

   The algorithm checks all items; use binary search on sorted list instead -> Option D

4. Quick Check:

   Linear search slow; binary search fast [OK]

Hint: Replace linear search with binary search for speed [OK]

Common Mistakes:

• Thinking syntax error exists
• Confusing memory use with speed
• Believing return value is wrong

5.

You have a list of 1,000,000 numbers sorted in order. You want to find if the number 500,000 is in the list. Which algorithm is best and why?

hard

A. Randomly pick numbers until you find 500,000

B. Check each number from start to end; simple but slow

C. Use binary search to jump and halve the search area repeatedly

D. Sort the list again before searching

Solution

1. Step 1: Understand the problem and data

   The list is sorted with 1,000,000 numbers; searching for 500,000.

2. Step 2: Evaluate algorithm choices

   Checking each number (Check each number from start to end; simple but slow) is slow. Random picking (Randomly pick numbers until you find 500,000) is unreliable. Sorting again (Sort the list again before searching) wastes time. Binary search (Use binary search to jump and halve the search area repeatedly) uses the sorted order to jump and halve search area, making it fastest.

3. Final Answer:

   Use binary search to jump and halve the search area repeatedly -> Option C

4. Quick Check:

   Sorted list + binary search = fastest search [OK]

Hint: Use binary search on sorted lists for fast lookup [OK]

Common Mistakes:

• Choosing linear search for large sorted lists
• Thinking sorting again helps
• Relying on random guessing