Why Two Sum Problem Classic Hash Solution in DSA Typescript?

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 your gift card amount. You try checking every pair manually by adding each price to every other price to see if they match your total.

The Problem

This manual way is slow and tiring because you have to check many pairs, especially if the list is long. It's easy to make mistakes or miss pairs, and it takes a lot of time to finish.

The Solution

The Two Sum problem with a hash solution uses a simple map to remember what numbers you've seen. For each number, you quickly check if the number needed to reach the target is already in the map. This way, you find the pair fast without checking every combination.

Before vs After

✗ Before

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

✓ After

const map = new Map<number, number>();
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);
}

What It Enables

This method lets you find pairs that add up to a target quickly and easily, even in very large lists.

Real Life Example

When shopping online, you want to quickly find two products that together use your full gift card without going over. This solution helps the website find those products instantly.

Key Takeaways

Manual pair checking is slow and error-prone.

Using a hash map remembers seen numbers for quick lookup.

Two Sum classic hash solution finds pairs fast and efficiently.

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