Why Two Sum Problem Classic Hash Solution in DSA Java?

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.

You try checking every possible pair manually to see if they add up to the gift card value.

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 what numbers you've seen.

It quickly finds if the matching number to reach the target exists without checking all pairs.

Before vs After

✗ Before

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

✓ After

Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
  int complement = target - nums[i];
  if (map.containsKey(complement)) return new int[]{map.get(complement), i};
  map.put(nums[i], i);
}

What It Enables

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

Real Life Example

When shopping online, you want to quickly find two products that together use your full budget without overspending.

Key Takeaways

Manual pair checking is slow and repeats work.

Using a hash map remembers seen numbers for quick matching.

Two Sum classic hash solution is efficient and easy to implement.

Practice

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

3. Final Answer:

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

4. Quick Check:

   Hash map = quick index lookup [OK]

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

Common Mistakes:

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

2. Which of the following is the correct Java syntax to declare a hash map for integer keys and values used in Two Sum?

easy

A. HashMap map = new HashMap();

B. HashMap map = new HashMap<>();

C. Map map = new HashMap<>();

D. HashMap map = new HashMap<>();

Solution

1. Step 1: Identify correct generic types for HashMap

   Java uses wrapper classes Integer, not primitive int, for generics.

2. Step 2: Check syntax correctness

   HashMap map = new HashMap<>(); correctly declares HashMap with diamond operator for instantiation.

3. Final Answer:

   HashMap map = new HashMap<>(); -> Option D

4. Quick Check:

   Use wrapper classes and diamond operator [OK]

Hint: Use Integer, not int, for HashMap generics in Java [OK]

Common Mistakes:

• Using primitive types int in generics
• Missing diamond operator <>
• Incorrect Map interface usage

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

public int[] twoSum(int[] nums, int target) {
    Map map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[] { map.get(complement), i };
        }
        map.put(nums[i], i);
    }
    return new int[] {};
}

medium

A. [1, 2]

B. [0, 1]

C. [2, 3]

D. []

Solution

1. Step 1: Trace the loop and map updates

   i=0: map={}, complement=9-2=7, not found, map={2:0} i=1: complement=9-7=2, found in map at index 0, return [0,1]

2. Step 2: Return indices of the two numbers that sum to target

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

3. Final Answer:

   [0, 1] -> Option B

4. Quick Check:

   2 + 7 = 9 [OK]

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

Common Mistakes:

• Returning indices in wrong order
• Not checking complement before insertion
• Returning empty array incorrectly

4. Identify the error in the following Two Sum Java method and select the fix:

public int[] twoSum(int[] nums, int target) {
    Map map = new HashMap<>();
    for (int i = 0; i <= nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[] { map.get(complement), i };
        }
        map.put(nums[i], i);
    }
    return new int[] {};
}

medium

A. Change loop condition to i < nums.length

B. Change map to TreeMap

C. Remove map.put line

D. Change return type to List<Integer>

Solution

1. Step 1: Identify the loop boundary error

   The loop uses i <= nums.length which causes ArrayIndexOutOfBoundsException at last iteration.

2. Step 2: Fix the loop condition

   Changing to i < nums.length prevents out-of-bound access and fixes the error.

3. Final Answer:

   Change loop condition to i < nums.length -> Option A

4. Quick Check:

   Loop index must be less than array length [OK]

Hint: Use i < length, not i <= length, for array loops [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Changing map type unnecessarily
• Removing essential map insertion

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

public int[] twoSum(int[] nums, int target) {
    Map map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[] { map.get(complement), i };
        }
        map.put(nums[i], i);
    }
    return new int[] {};
}

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, not found, map={3:0} i=1: complement=6-3=3, found at index 0, return [0,1]

2. Step 2: Understand handling of duplicates

   Even with duplicates, the first pair found is indices 0 and 1, both 3s summing to 6.

3. Final Answer:

   [0, 1] -> Option A

4. Quick Check:

   3 + 3 = 6 [OK]

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

Common Mistakes:

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