Why Two Sum Problem Classic Hash Solution in DSA C++?

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 Classic Hash Solution uses a fast lookup method to remember what numbers you've seen.

This way, you can quickly find if the matching number to reach your target exists without checking all pairs.

Before vs After

✗ Before

for (int i = 0; i < nums.size(); i++) {
  for (int j = i + 1; j < nums.size(); j++) {
    if (nums[i] + nums[j] == target) return {i, j};
  }
}

✓ After

unordered_map<int, int> seen;
for (int i = 0; i < nums.size(); i++) {
  int complement = target - nums[i];
  if (seen.count(complement)) return {seen[complement], i};
  seen[nums[i]] = 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, the system can instantly suggest two products that fit your budget without waiting long.

Key Takeaways

Manual pair checking is slow and inefficient.

Using a hash map speeds up finding matching pairs.

This approach works well for large data sets and real-time applications.

Practice

1. What is the main purpose of using a hash map in the classic Two Sum problem solution?

easy

A. To store numbers and their indices for quick lookup

B. To sort the array before searching

C. To store the sum of all pairs

D. To reverse the array elements

Solution

1. Step 1: Understand the role of hash map

   The hash map stores each number as a key and its index as the value for quick access.

2. Step 2: Explain why quick lookup is needed

   Quick lookup helps find if the complement (target - current number) exists without scanning the entire array.

3. Final Answer:

   To store numbers and their indices for quick lookup -> Option A

4. Quick Check:

   Hash map = quick index lookup [OK]

Hint: Hash map stores values for instant complement search [OK]

Common Mistakes:

• Thinking hash map sorts the array
• Confusing hash map with storing sums
• Assuming hash map reverses elements

2. Which of the following C++ code snippets correctly declares the hash map used in the Two Sum problem to store integer keys and values?

easy

A. std::map<int> map;

B. std::unordered_map map;

C. std::unordered_map<int> map;

D. std::vector map;

Solution

1. Step 1: Identify correct hash map syntax

   std::unordered_map declares a hash map with int keys and int values.

2. Step 2: Explain why others are incorrect

   std::map<int> lacks value type; std::unordered_map<int> lacks value type; std::vector is invalid syntax.

3. Final Answer:

   std::unordered_map<int, int> map; -> Option B

4. Quick Check:

   Correct hash map syntax = std::unordered_map<int, int> [OK]

Hint: Use std::unordered_map<key, value> for hash maps in C++ [OK]

Common Mistakes:

• Omitting value type in map declaration
• Using std::vector instead of map
• Confusing std::map and std::unordered_map syntax

3. What will be the output of the following C++ code snippet for the input array nums = {2, 7, 11, 15} and target = 9?

std::vector<int> twoSum(std::vector<int>& nums, int target) {
    std::unordered_map<int, int> map;
    for (int i = 0; i < nums.size(); i++) {
        int complement = target - nums[i];
        if (map.find(complement) != map.end()) {
            return {map[complement], i};
        }
        map[nums[i]] = i;
    }
    return {};
}

int main() {
    auto result = twoSum({2, 7, 11, 15}, 9);
    for (int i : result) std::cout << i << " ";
}

medium

A. [1 0]

B. [2 3]

C. [0 1]

D. No output

Solution

1. Step 1: Trace the loop and map updates

   At i=0, map is empty, insert (2,0). At i=1, complement=9-7=2 found in map at index 0.

2. Step 2: Return indices of complement and current

   Return {0,1} which prints as "0 1".

3. Final Answer:

   [0 1] -> Option C

4. Quick Check:

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

Hint: Check complement in map before inserting current number [OK]

Common Mistakes:

• Returning indices in wrong order
• Not checking complement before insertion
• Confusing values with indices

4. Identify the bug in this Two Sum function and select the fix:

std::vector<int> twoSum(std::vector<int>& nums, int target) {
    std::unordered_map<int, int> map;
    for (int i = 0; i < nums.size(); i++) {
        int complement = target - nums[i];
        if (map.find(complement) != map.end()) {
            return {i, map[complement]};
        }
        map[nums[i]] = i;
    }
    return {};
}

medium

A. Swap the return indices to {map[complement], i}

B. Change unordered_map to map

C. Insert nums[i] into map before checking complement

D. Return empty vector immediately

Solution

1. Step 1: Analyze the return statement order

   The function returns {i, map[complement]} but correct order is {map[complement], i} to match indices.

2. Step 2: Explain why swapping fixes the bug

   Swapping ensures the first index is of complement and second is current index, matching problem requirement.

3. Final Answer:

   Swap the return indices to {map[complement], i} -> Option A

4. Quick Check:

   Correct index order = complement index, current index [OK]

Hint: Return complement index first, then current index [OK]

Common Mistakes:

• Returning indices in wrong order
• Inserting before checking complement
• Changing map type unnecessarily

5. Given the input array nums = {3, 3, 4, 5} and target = 6, what will the classic Two Sum hash solution return and why?

std::vector<int> twoSum(std::vector<int>& nums, int target) {
    std::unordered_map<int, int> map;
    for (int i = 0; i < nums.size(); i++) {
        int complement = target - nums[i];
        if (map.find(complement) != map.end()) {
            return {map[complement], i};
        }
        map[nums[i]] = i;
    }
    return {};
}

hard

A. Empty vector because no two numbers sum to 6

B. [2 3] because 4 and 5 sum to 9, not 6

C. [1 0] because indices are reversed

D. [0 1] because first 3 and second 3 sum to 6

Solution

1. Step 1: Trace the loop with duplicates

   At i=0, map empty, insert (3,0). At i=1, complement=6-3=3 found at index 0.

2. Step 2: Return indices of first and second 3

   Return {0,1} because nums[0] + nums[1] = 3 + 3 = 6.

3. Final Answer:

   [0 1] because first 3 and second 3 sum to 6 -> Option D

4. Quick Check:

   Duplicates allowed, indices 0 and 1 sum to target [OK]

Hint: Duplicates can form valid pairs if sum matches target [OK]

Common Mistakes:

• Ignoring duplicates in map
• Assuming no pair sums to target
• Returning indices in wrong order