Two Sum Problem Classic Hash Solution in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the dictionary for storing seen numbers.

DSA Python

seen = [1]

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

A[]

B{}

C()

Dset()

Attempts:

3 left

💡 Hint

Common Mistakes

Using a list or set instead of a dictionary.

Using parentheses which create a tuple, not a dictionary.

We use an empty dictionary {} to store numbers and their indices for quick lookup.

2fill in blank

medium

Complete the code to iterate over the list with index and value.

DSA Python

for [1], num in enumerate(nums):

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

Ai

Bvalue

Cindex

Ditem

Attempts:

3 left

💡 Hint

Common Mistakes

Using the value variable name instead of index.

Using a variable name that is not descriptive or inconsistent.

The variable i is commonly used as the index in loops.

3fill in blank

hard

Fix the error in checking if the complement exists in the dictionary.

DSA Python

if [1] in seen:

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

Anum

Bi

Ctarget - num

Dseen[num]

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if the current number num is in seen instead of the complement.

Using the index i instead of the complement.

We check if the complement target - num is already in the dictionary seen.

4fill in blank

hard

Fill both blanks to return the indices of the two numbers that add up to target.

DSA Python

return [seen[[1]], [2]]

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

Atarget - num

Bi

Cnum

Dseen[num]

Attempts:

3 left

💡 Hint

Common Mistakes

Returning the current number instead of its index.

Swapping the order of indices incorrectly.

Return the index of the complement from seen and the current index i.

5fill in blank

hard

Fill all three blanks to add the current number and its index to the dictionary.

DSA Python

seen[[1]] = [2]

# After loop ends
return [3]

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

Anum

Bi

C[]

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Using the index as key instead of the number.

Returning an empty list instead of None.

Add the current number as key and index as value to seen. Return None if no pair found.

Practice

1. What is the main idea behind the classic hash solution for the Two Sum Problem?

easy

A. Use a dictionary to store numbers and their indices for quick lookup

B. Sort the array and use two pointers to find the sum

C. Use nested loops to check every pair of numbers

D. Use recursion to find pairs that add up to the target

Solution

1. Step 1: Understand the problem requirement

   The Two Sum problem asks to find two numbers in an array that add up to a target value.

2. Step 2: Explain the hash solution approach

   Using a dictionary (hash map) allows storing each number's index and checking if the complement (target - current number) exists quickly.

3. Final Answer:

   Use a dictionary to store numbers and their indices for quick lookup -> Option A

4. Quick Check:

   Hash map lookup = fast pair finding [OK]

Hint: Use a dictionary to find complements quickly [OK]

Common Mistakes:

• Trying to sort the array which changes indices
• Using nested loops causing slow performance
• Not storing indices, only values

2. Which of the following is the correct syntax to check if a key exists in a Python dictionary named seen?

easy

A. if key exists in seen:

B. if key in seen:

C. if seen.has_key(key):

D. if key inside seen:

Solution

1. Step 1: Recall Python dictionary membership syntax

   In Python, to check if a key exists in a dictionary, use the syntax if key in dict:.

2. Step 2: Identify correct option

   if key in seen: uses the correct syntax. if key exists in seen: and D are invalid syntax. if seen.has_key(key): is legacy and removed in Python 3.

3. Final Answer:

   if key in seen: -> Option B

4. Quick Check:

   Python dict key check = if key in dict [OK]

Hint: Use 'in' keyword to check dictionary keys [OK]

Common Mistakes:

• Using outdated has_key() method
• Writing invalid syntax like 'exists in'
• Confusing keys with values

3. What will be the output of the following code?

def two_sum(nums, target):
    seen = {}
    for i, num in enumerate(nums):
        complement = target - num
        if complement in seen:
            return [seen[complement], i]
        seen[num] = i
    return []

print(two_sum([2, 7, 11, 15], 9))

medium

A. [0, 1]

B. [1, 2]

C. [2, 3]

D. []

Solution

1. Step 1: Trace the loop and dictionary updates

   Iteration 0: num=2, complement=7, seen={}, complement not in seen, add 2:0.
   Iteration 1: num=7, complement=2, complement 2 in seen at index 0, return [0,1].

2. Step 2: Determine returned indices

   The function returns indices [0,1] where nums[0]+nums[1]=2+7=9.

3. Final Answer:

   [0, 1] -> Option A

4. Quick Check:

   Indices of 2 and 7 = [0, 1] [OK]

Hint: Check complement in dict before adding current number [OK]

Common Mistakes:

• Returning values instead of indices
• Adding current number before checking complement
• Returning empty list incorrectly

4. The following code tries to solve the Two Sum problem but has a bug. What is the bug?

def two_sum(nums, target):
    seen = {}
    for i, num in enumerate(nums):
        if num in seen:
            return [seen[num], i]
        complement = target - num
        seen[complement] = i
    return []

medium

A. It does not check for complement in dictionary

B. It returns indices in wrong order

C. It stores complement as key instead of the number itself

D. It uses enumerate incorrectly

Solution

1. Step 1: Analyze dictionary key storage

   The code stores complement as key with current index, but it should store the number as key to check complement later.

2. Step 2: Identify impact of wrong key storage

   Because complement is stored as key, the lookup for num in seen fails to find correct pairs, causing wrong or no results.

3. Final Answer:

   It stores complement as key instead of the number itself -> Option C

4. Quick Check:

   Store number as key, not complement [OK]

Hint: Store numbers as keys, check for complement [OK]

Common Mistakes:

• Swapping complement and number keys
• Returning indices before checking complement
• Not using enumerate properly

5. Given the list nums = [3, 3, 4, 5] and target = 6, what will the classic hash solution return?
Consider the code:

def two_sum(nums, target):
    seen = {}
    for i, num in enumerate(nums):
        complement = target - num
        if complement in seen:
            return [seen[complement], i]
        seen[num] = i
    return []

print(two_sum(nums, target))

hard

A. []

B. [1, 2]

C. [2, 3]

D. [0, 1]

Solution

1. Step 1: Trace the loop with duplicates

   Iteration 0: num=3, complement=3, seen={}, complement not in seen, add 3:0.
   Iteration 1: num=3, complement=3, complement 3 in seen at index 0, return [0,1].

2. Step 2: Understand duplicate handling

   The solution correctly finds the two 3's at indices 0 and 1 that sum to 6.

3. Final Answer:

   [0, 1] -> Option D

4. Quick Check:

   Two 3's indices = [0, 1] [OK]

Hint: Duplicates allowed; first pair found is returned [OK]

Common Mistakes:

• Ignoring duplicates and returning empty list
• Returning wrong indices due to overwriting
• Assuming no pair exists when duplicates present