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 a gift card amount. You try to check every pair manually to see if they match the total.

The Problem

Checking every pair one by one takes a lot of time, especially if the list is long. It's easy to 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 numbers seen before. It quickly finds if the matching number exists without checking all pairs, saving time and effort.

Before vs After

✗ Before

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

✓ After

for (int i = 0; i < n; i++) {
  int complement = target - nums[i];
  if (hash_contains(complement)) return (hash_get(complement), i);
  hash_put(nums[i], i);
}

What It Enables

This method makes finding pairs that add up to a target fast and easy, even with large lists.

Real Life Example

When shopping online, quickly finding two products that fit your budget without checking every combination saves time and helps you decide faster.

Key Takeaways

Manual pair checking is slow and error-prone.

Hashing remembers seen numbers for quick lookup.

Two Sum Classic Hash Solution finds pairs efficiently.

Practice

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

easy

A. To store all pairs of numbers

B. To sort the array before searching

C. To store numbers and their indices for quick lookup

D. To reverse the array elements

Solution

1. Step 1: Understand the role of hash map in Two Sum

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

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 again.

3. Final Answer:

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

4. Quick Check:

   Hash map = quick lookup [OK]

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

Common Mistakes:

• Thinking hash map sorts the array
• Assuming hash map stores all pairs
• Confusing hash map with array reversal

2. Which of the following is the correct syntax to declare a hash map (dictionary) in C for the Two Sum problem?

easy

A. unordered_map hashmap;

B. int hashmap[1000];

C. map hashmap = new map();

D. dict hashmap = {};

Solution

1. Step 1: Identify C++ STL syntax for hash map

   In C++, unordered_map is used for hash maps with integer keys and values.

2. Step 2: Check other options for correctness

   int hashmap[1000]; is a fixed-size array, not a hash map. map hashmap = new map(); is invalid syntax. dict hashmap = {}; is Python syntax.

3. Final Answer:

   unordered_map hashmap; -> Option A

4. Quick Check:

   unordered_map = hash map in C++ [OK]

Hint: Use unordered_map for hash map in C++ [OK]

Common Mistakes:

• Using array instead of hash map
• Confusing Python or Java syntax with C++
• Incorrect map declaration syntax

3. What will be the output of the following code snippet for input nums = [2, 7, 11, 15], target = 9?

int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
    unordered_map map;
    for (int i = 0; i < numsSize; i++) {
        int complement = target - nums[i];
        if (map.find(complement) != map.end()) {
            int* result = (int*)malloc(2 * sizeof(int));
            result[0] = map[complement];
            result[1] = i;
            *returnSize = 2;
            return result;
        }
        map[nums[i]] = i;
    }
    *returnSize = 0;
    return NULL;
}

medium

A. [1, 0]

B. [0, 1]

C. [2, 3]

D. NULL

Solution

1. Step 1: Trace the loop and hash map insertions

   At i=0, map = {2:0}. At i=1, complement = 9-7=2 found in map at index 0.

2. Step 2: Return indices of complement and current number

   Return [0,1] as the pair that sums to 9.

3. Final Answer:

   [0, 1] -> Option B

4. Quick Check:

   nums[0]+nums[1] = 2+7 = 9 [OK]

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

Common Mistakes:

• Returning indices in wrong order
• Not checking complement before insertion
• Returning NULL when pair exists

4. Identify the bug in this Two Sum hash map solution snippet:

for (int i = 0; i < numsSize; i++) {
    map[nums[i]] = i;
    int complement = target - nums[i];
    if (map.find(complement) != map.end()) {
        return {map[complement], i};
    }
}

medium

A. Loop should start from i = 1

B. map should store complement instead of current number

C. Return statement syntax is invalid in C

D. Complement check should happen before inserting current number

Solution

1. Step 1: Analyze order of operations in loop

   Current number is inserted before checking complement, causing false positive when complement equals current number.

2. Step 2: Correct order to check complement before insertion

   Check if complement exists in map first, then insert current number to avoid matching with itself.

3. Final Answer:

   Complement check should happen before inserting current number -> Option D

4. Quick Check:

   Check complement before insert [OK]

Hint: Check complement before adding current number to map [OK]

Common Mistakes:

• Inserting before checking complement
• Wrong return syntax in C
• Starting loop from wrong index

5. Given an array nums = [3, 3, 4, 5] and target = 6, what will the Two Sum hash map solution return and why?

hard

A. [0, 1] because nums[0] + nums[1] = 3 + 3 = 6

B. [2, 3] because nums[2] + nums[3] = 4 + 5 = 9

C. NULL because no two numbers sum to 6

D. [1, 0] because indices order is reversed

Solution

1. Step 1: Trace the hash map insertions and complement checks

   At i=0, map = {3:0}. At i=1, complement = 6-3=3 found at index 0.

2. Step 2: Return indices of the first valid pair

   Return [0,1] as the pair that sums to 6.

3. Final Answer:

   [0, 1] because nums[0] + nums[1] = 3 + 3 = 6 -> Option A

4. Quick Check:

   3 + 3 = 6 [OK]

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

Common Mistakes:

• Ignoring duplicate values
• Returning wrong indices order
• Assuming no pair exists when duplicates present