Two Sum Problem Classic Hash Solution in DSA C++ - Time & Space Complexity

Time Complexity: Two Sum Problem Classic Hash Solution

O(n)

Understanding Time Complexity

We want to understand how fast the classic hash-based solution for the Two Sum problem runs.

Specifically, how the time to find two numbers that add up to a target grows as the input list gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <unordered_map>
#include <vector>

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

This code finds two numbers in the list that add up to the target using a hash map for quick lookups.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop over the input array.
• How many times: Exactly once for each element in the array.
• Inside the loop: A constant-time hash map lookup and insertion.

How Execution Grows With Input

As the input size grows, the number of loop iterations grows linearly.

Input Size (n)	Approx. Operations
10	About 10 lookups and insertions
100	About 100 lookups and insertions
1000	About 1000 lookups and insertions

Pattern observation: The operations increase directly with input size, doubling input roughly doubles work.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the two numbers grows in a straight line with the number of elements.

Common Mistake

[X] Wrong: "Because there is a loop inside a loop, the time is O(n²)."

[OK] Correct: There is only one loop; the hash map lookups are fast and do not add extra loops.

Interview Connect

Understanding this solution's time complexity shows you can use data structures smartly to speed up searches.

This skill is valuable for solving many problems efficiently in interviews and real projects.

Self-Check

"What if the input array was sorted and we used two pointers instead of a hash map? How would the time complexity change?"

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