Why Two Sum Problem Classic Hash Solution in DSA Go?

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 together cost exactly your gift card amount. Doing this by checking every pair one by one is like trying to find a friend in a crowd by asking each person individually--it takes a lot of time and effort.

The Problem

Checking every possible pair manually means you spend a lot of time, especially if the list is long. It's easy to make mistakes or miss pairs because you have to remember which items you already checked. This slow and error-prone way makes you frustrated and wastes your time.

The Solution

The classic hash solution uses a simple map to remember what numbers you've seen. Instead of checking all pairs, you look up if the number needed to reach your target is already in the map. This way, you find the answer quickly and without confusion, like having a friend who instantly tells you if your match is nearby.

Before vs After

✗ Before

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

✓ After

seen := make(map[int]int)
for i, num := range nums {
    complement := target - num
    if j, found := seen[complement]; found {
        return []int{j, i}
    }
    seen[num] = i
}

What It Enables

This method lets you find pairs that add up to a target instantly, even in huge lists, saving time and effort.

Real Life Example

When shopping online, you want to quickly find two products that fit your budget exactly without trying every combination manually.

Key Takeaways

Manual pair checking is slow and error-prone.

Using a hash map remembers seen numbers for quick lookup.

This approach finds pairs in one pass, making it fast and simple.

Practice

1. What is the main idea behind the Two Sum Problem Classic Hash Solution?

easy

A. Use a hash map to store numbers and their indices for quick lookup

B. Sort the array and use two pointers to find the sum

C. Use nested loops to check every pair of numbers

D. Use recursion to find pairs that add up to the target

Solution

1. Step 1: Understand the problem goal

   The goal is to find two numbers in the array that add up to a target value.

2. Step 2: Use a hash map for efficient lookup

   By storing each number and its index in a hash map, we can check if the complement (target - current number) exists quickly.

3. Final Answer:

   Use a hash map to store numbers and their indices for quick lookup -> Option A

4. Quick Check:

   Hash map lookup = fast pair finding [OK]

Hint: Use a hash map to find complements quickly [OK]

Common Mistakes:

• Trying to use nested loops which is slower
• Sorting the array which is unnecessary here
• Not storing indices with numbers

2. Which of the following Go code snippets correctly initializes a hash map to store integers as keys and values?

easy

A. var m map[int]int

B. m := map[int]int

C. m := make(map[int]int)

D. m := map[int]int(make())

Solution

1. Step 1: Recall Go map initialization syntax

   Maps in Go are initialized using the built-in make function like make(map[keyType]valueType).

2. Step 2: Identify correct syntax

   m := make(map[int]int) uses make(map[int]int) which is the correct way to create an empty map with int keys and int values.

3. Final Answer:

   m := make(map[int]int) -> Option C

4. Quick Check:

   make(map[int]int) = correct map init [OK]

Hint: Use make(map[key]value) to create maps in Go [OK]

Common Mistakes:

• Using var without make causes nil map
• Incorrect syntax like map[int]int(make())
• Using empty braces without make in some contexts

3. What will be the output of the following Go code snippet?

func twoSum(nums []int, target int) []int {
    m := make(map[int]int)
    for i, num := range nums {
        if j, ok := m[target-num]; ok {
            return []int{j, i}
        }
        m[num] = i
    }
    return nil
}

fmt.Println(twoSum([]int{2, 7, 11, 15}, 9))

medium

A. [1 0]

B. [0 1]

C. [2 3]

D. nil

Solution

1. Step 1: Trace the loop and map updates

   At i=0, num=2, map is empty, so store m[2]=0.
   At i=1, num=7, check if m[9-7=2] exists, yes at index 0.
   Return [0,1].

2. Step 2: Confirm returned indices

   The function returns indices of numbers 2 and 7 which sum to 9, so output is [0 1].

3. Final Answer:

   [0 1] -> Option B

4. Quick Check:

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

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

Common Mistakes:

• Returning indices in wrong order
• Not checking map before insertion
• Returning nil when pair exists

4. Identify the error in the following Go code for the Two Sum problem:

func twoSum(nums []int, target int) []int {
    m := make(map[int]int)
    for i, num := range nums {
        if j, ok := m[num-target]; ok {
            return []int{j, i}
        }
        m[num] = i
    }
    return nil
}

medium

A. The complement calculation should be target - num, not num - target

B. The map should store num + target as key

C. The map initialization is incorrect

D. The return statement should return nil earlier

Solution

1. Step 1: Analyze complement calculation

   The complement to find is target - current number, not num - target, because we want a number that adds with num to target.

2. Step 2: Identify the fix

   Change m[num-target] to m[target-num] to correctly check for the complement in the map.

3. Final Answer:

   The complement calculation should be target - num, not num - target -> Option A

4. Quick Check:

   Complement = target - num [OK]

Hint: Complement = target minus current number [OK]

Common Mistakes:

• Swapping target and num in complement
• Incorrect map initialization
• Returning nil too early

5. Given the input array nums = [3, 3, 4, 2] and target 6, what will the Two Sum Classic Hash Solution return?
Consider the code below:

func twoSum(nums []int, target int) []int {
    m := make(map[int]int)
    for i, num := range nums {
        if j, ok := m[target-num]; ok {
            return []int{j, i}
        }
        m[num] = i
    }
    return nil
}

medium

A. nil

B. [1 2]

C. [2 3]

D. [0 1]

Solution

1. Step 1: Trace the loop with given input

   i=0, num=3, map empty, store m[3]=0.
   i=1, num=3, check m[6-3=3], found at index 0, return [0,1].

2. Step 2: Confirm returned indices

   Indices 0 and 1 correspond to values 3 and 3 which sum to 6, so output is [0 1].

3. Final Answer:

   [0 1] -> Option D

4. Quick Check:

   3 + 3 = 6 at indices [0 1] [OK]

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

Common Mistakes:

• Assuming different pairs without checking map
• Returning nil when pair exists
• Confusing indices order