Why Two Sum Problem Classic Hash Solution in DSA Python?

The Big Idea

What if you could find matching pairs instantly without checking every possibility?

The Scenario

Imagine you have a list of prices for items in a store, and you want to find two items that add up to exactly the amount of money you have in your wallet.

You try checking every possible pair one by one to see if their prices add up to your money.

The Problem

Checking every pair manually takes a lot of time, especially if the list is long.

You might miss pairs or repeat checks, making it slow and frustrating.

The Solution

The Two Sum problem classic hash solution uses a simple helper (a hash map) to remember what prices you've seen.

This way, you can quickly check if the partner price needed to reach your total is already seen, making the search very fast and easy.

Before vs After

✗ Before

for i in range(len(nums)):
    for j in range(i+1, len(nums)):
        if nums[i] + nums[j] == target:
            return [i, j]

✓ After

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

What It Enables

This method lets you find pairs that add up to a target quickly, even in very large lists.

Real Life Example

When shopping online, you want to quickly find two products that fit your budget without trying every combination.

Key Takeaways

Manual pair checking is slow and repeats work.

Using a hash map remembers seen numbers for quick lookup.

This makes finding pairs fast and efficient.

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