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

Time Complexity: Two Sum Problem Classic Hash Solution

O(n)

Understanding Time Complexity

We want to understand how the time needed to find two numbers that add up to a target grows as the list gets bigger.

How does the number of steps change when the input list size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int[] twoSum(int[] nums, int target) {
    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);
    }
    return new int[0];
}

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: Single loop through the array.
• How many times: Exactly once for each element in the array.
• Inside the loop, checking and inserting in the hash map happens in constant time.

How Execution Grows With Input

As the list gets bigger, the number of steps grows directly with the number of elements.

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 steps increase evenly as the input size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the two numbers grows in a straight line with the size of the input list.

Common Mistake

[X] Wrong: "Since 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 shows you can use extra space to speed up searching, a common and useful skill in coding problems.

Self-Check

"What if the input array was already sorted and you 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 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