Two Sum Problem Classic Hash Solution in DSA Python - Build from Scratch

Two Sum Problem Classic Hash Solution

📖 Scenario: You are helping a friend who runs a small store. They want to find two items that together cost exactly a certain amount of money. You will write a program to find the positions of these two items in the price list.

🎯 Goal: Build a program that finds two numbers in a list that add up to a target value using a hash map (dictionary) for fast lookup.

📋 What You'll Learn

Create a list called prices with the exact values: [2, 7, 11, 15]

Create a variable called target and set it to 9

Use a dictionary called seen to store numbers and their indices while looping through prices

Find two indices i and j such that prices[i] + prices[j] == target

Print the list of the two indices as the final output

💡 Why This Matters

🌍 Real World

Finding two items in a list that add up to a specific value is useful in shopping apps, budgeting tools, and financial software.

💼 Career

This problem is a common coding interview question and helps develop skills in using hash maps for efficient searching.

1

Create the list of prices

Create a list called prices with these exact values: [2, 7, 11, 15]

DSA Python

# Create the list called prices with values [2, 7, 11, 15]
# Your code here

Hint

Use square brackets to create a list and separate numbers with commas.

2

Set the target sum

Create a variable called target and set it to 9

DSA Python

prices = [2, 7, 11, 15]
# Create the variable target and set it to 9
# Your code here

Hint

Use the equals sign to assign the value 9 to the variable target.

3

Find two indices using a dictionary

Create an empty dictionary called seen. Use a for loop with variables i and num to iterate over enumerate(prices). Inside the loop, check if target - num is in seen. If yes, create a list result with seen[target - num] and i, then break the loop. Otherwise, add num as a key and i as its value to seen.

DSA Python

prices = [2, 7, 11, 15]
target = 9
# Create dictionary seen and use a for loop to find two indices
# Your code here

Hint

Use a dictionary to remember numbers and their positions. Check if the complement exists before adding the current number.

4

Print the result

Print the variable result to show the two indices found.

DSA Python

prices = [2, 7, 11, 15]
target = 9
seen = {}
for i, num in enumerate(prices):
    if target - num in seen:
        result = [seen[target - num], i]
        break
    seen[num] = i
# Print the result list
# Your code here

Hint

Use the print function to display the list of indices.

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