Two Sum Problem Classic Hash Solution in DSA Typescript - 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 as the input grows.

Specifically, how does the number of steps change when the list of numbers gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function twoSum(nums: number[], target: number): number[] {
  const map = new Map();
  for (let i = 0; i < nums.length; i++) {
    const complement = target - nums[i];
    if (map.has(complement)) {
      return [map.get(complement)!, i];
    }
    map.set(nums[i], i);
  }
  return [];
}
    

This code finds two numbers in the array 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: One loop that goes through each number once.
• How many times: Exactly n times, where n is the length of the input array.
• Inside the loop, each hash map check and insert is done in constant time.

How Execution Grows With Input

As the input size grows, the number of steps grows roughly the same way.

Input Size (n)	Approx. Operations
10	About 10 checks and inserts
100	About 100 checks and inserts
1000	About 1000 checks and inserts

Pattern observation: The work grows linearly with input size; doubling input roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time it takes grows in direct proportion to the number of elements in the input.

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 operations inside the loop are fast and do not add extra loops, so the time stays linear.

Interview Connect

Understanding this solution's time complexity shows you can use data structures smartly to speed up problems that might seem slow at first.

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 Two Sum problem classic solution?

easy

A. To store the sum of all pairs

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

   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 to 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 stores number:index = A [OK]

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

Common Mistakes:

• Thinking hash map sorts the array
• Trying to store sums instead of indices
• Using hash map to reverse elements

2. Which of the following TypeScript code snippets correctly initializes an empty hash map for the Two Sum problem?

easy

A. const map = new Map();

B. const map = {};

C. const map = new Array();

D. const map = new Set();

Solution

1. Step 1: Identify correct hash map initialization

   In TypeScript, new Map<number, number>() creates a map with number keys and values.

2. Step 2: Explain why other options are incorrect

   {} is an object but lacks type safety; new Array() is an array, not a map; new Set() stores unique values, not key-value pairs.

3. Final Answer:

   const map = new Map<number, number>(); -> Option A

4. Quick Check:

   Use Map<number, number> for hash map [OK]

Hint: Use Map<number, number> for key-value pairs [OK]

Common Mistakes:

• Using Set instead of Map
• Using Array instead of Map
• Using plain object without type safety

3. What will be the output of the following TypeScript code for the input array [2, 7, 11, 15] and target 9?

function twoSum(nums: number[], target: number): number[] {
  const map = new Map();
  for (let i = 0; i < nums.length; i++) {
    const complement = target - nums[i];
    if (map.has(complement)) {
      return [map.get(complement)!, i];
    }
    map.set(nums[i], i);
  }
  return [];
}

console.log(twoSum([2, 7, 11, 15], 9));

medium

A. [1, 2]

B. []

C. [2, 3]

D. [0, 1]

Solution

1. Step 1: Trace the loop and map updates

   i=0: map={}, complement=9-2=7, map.has(7)? No, map.set(2,0).
   i=1: complement=9-7=2, map.has(2)? Yes, index=0, return [0,1].

2. Step 2: Confirm returned indices

   Indices 0 and 1 correspond to numbers 2 and 7 which sum to 9.

3. Final Answer:

   [0, 1] -> Option D

4. Quick Check:

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

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

Common Mistakes:

• Returning values instead of indices
• Not checking map before insertion
• Returning empty array incorrectly

4. Identify the bug in this TypeScript Two Sum implementation:

function twoSum(nums: number[], target: number): number[] {
  const map = new Map();
  for (let i = 0; i <= nums.length; i++) {
    const complement = target - nums[i];
    if (map.has(complement)) {
      return [map.get(complement)!, i];
    }
    map.set(nums[i], i);
  }
  return [];
}

medium

A. map.set should be called before checking complement

B. Loop condition should be < instead of <=

C. Return statement should return values instead of indices

D. Complement calculation is incorrect

Solution

1. Step 1: Analyze loop condition

   The loop uses i <= nums.length, which causes out-of-bounds access at last iteration.

2. Step 2: Correct loop condition

   Change to i < nums.length to avoid accessing undefined element.

3. Final Answer:

   Loop condition should be < instead of <= -> Option B

4. Quick Check:

   Loop must stay within array bounds [OK]

Hint: Use < not <= for array index loops [OK]

Common Mistakes:

• Off-by-one error in loop condition
• Swapping order of map.set and map.has
• Returning values instead of indices

5. Given the input array [3, 3, 4, 5] and target 6, what will the Two Sum classic hash solution return?

function twoSum(nums: number[], target: number): number[] {
  const map = new Map();
  for (let i = 0; i < nums.length; i++) {
    const complement = target - nums[i];
    if (map.has(complement)) {
      return [map.get(complement)!, i];
    }
    map.set(nums[i], i);
  }
  return [];
}

console.log(twoSum([3, 3, 4, 5], 6));

hard

A. [0, 1]

B. [1, 2]

C. [2, 3]

D. []

Solution

1. Step 1: Trace the loop and map updates with duplicates

   i=0: map={}, complement=6-3=3, map.has(3)? No, map.set(3,0).
   i=1: complement=6-3=3, map.has(3)? Yes, index=0, return [0,1].

2. Step 2: Confirm correct pair with duplicates

   Even with duplicates, first pair indices 0 and 1 sum to 6.

3. Final Answer:

   [0, 1] -> Option A

4. Quick Check:

   Duplicates handled, indices [0,1] sum to target [OK]

Hint: Duplicates allowed; first matching pair returned [OK]

Common Mistakes:

• Ignoring duplicates and returning wrong indices
• Returning empty array incorrectly
• Confusing values with indices