Two Sum IV in O(n) Time Complexity using Java

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input will have at most one solution, and you may not use the same index twice.

In case no solution exists, return [-1, -1]

Example:

Input: nums = [2, 7, 11, 15], target = 9
Output: [0, 1]
Explanation: nums[0] + nums[1] = 2 + 7 = 9

Note:

Your algorithm should run in O(n) time and use O(n) space.

Understanding the Problem

The core challenge of this problem is to find two indices in an array such that the elements at these indices sum up to a given target. This problem is significant in various applications, such as financial transactions, where you need to find pairs of transactions that sum up to a specific amount.

Potential pitfalls include not considering that the same index cannot be used twice and ensuring that the solution is efficient in terms of time complexity.

Approach

To solve this problem, we can start with a naive approach and then optimize it:

Naive Solution

The naive solution involves using two nested loops to check all possible pairs of indices. This approach has a time complexity of O(n^2), which is not efficient for large arrays.

Optimized Solution

To achieve an O(n) time complexity, we can use a HashMap to store the elements and their indices as we iterate through the array. For each element, we check if the complement (target - current element) exists in the HashMap. If it does, we have found our solution.

Algorithm

Here is a step-by-step breakdown of the optimized algorithm:

Initialize an empty HashMap to store elements and their indices.
Iterate through the array.
For each element, calculate its complement (target - current element).
Check if the complement exists in the HashMap.
If it exists, return the indices of the current element and the complement.
If it does not exist, add the current element and its index to the HashMap.
If no solution is found, return [-1, -1].

Code Implementation

import java.util.HashMap;

public class TwoSum {
    public int[] twoSum(int[] nums, int target) {
        // Create a HashMap to store the value and its index
        HashMap<Integer, Integer> map = new HashMap<>();
        
        // Iterate through the array
        for (int i = 0; i < nums.length; i++) {
            // Calculate the complement
            int complement = target - nums[i];
            
            // Check if the complement exists in the map
            if (map.containsKey(complement)) {
                // If it exists, return the indices
                return new int[] { map.get(complement), i };
            }
            
            // If it does not exist, add the current element and its index to the map
            map.put(nums[i], i);
        }
        
        // If no solution is found, return [-1, -1]
        return new int[] { -1, -1 };
    }
}

Complexity Analysis

The time complexity of this approach is O(n) because we only iterate through the array once. The space complexity is also O(n) due to the HashMap storing up to n elements.

Edge Cases

Potential edge cases include:

An empty array: The function should return [-1, -1].
An array with no valid pairs: The function should return [-1, -1].
Negative numbers and zero: The function should handle these correctly.

Testing

To test the solution comprehensively, consider the following test cases:

public static void main(String[] args) {
    TwoSum ts = new TwoSum();
    
    // Test case 1: Normal case
    int[] result1 = ts.twoSum(new int[] {2, 7, 11, 15}, 9);
    System.out.println(Arrays.toString(result1)); // Expected: [0, 1]
    
    // Test case 2: No solution
    int[] result2 = ts.twoSum(new int[] {1, 2, 3}, 7);
    System.out.println(Arrays.toString(result2)); // Expected: [-1, -1]
    
    // Test case 3: Negative numbers
    int[] result3 = ts.twoSum(new int[] {-1, -2, -3, -4, -5}, -8);
    System.out.println(Arrays.toString(result3)); // Expected: [2, 4]
    
    // Test case 4: Zero and negative numbers
    int[] result4 = ts.twoSum(new int[] {0, 4, 3, 0}, 0);
    System.out.println(Arrays.toString(result4)); // Expected: [0, 3]
}

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Understand the problem requirements and constraints thoroughly.
Start with a brute force solution to understand the basic logic.
Think about how to optimize the solution using data structures like HashMaps.
Consider edge cases and test your solution comprehensively.

Conclusion

In this blog post, we discussed the Two Sum problem and provided an optimized solution using a HashMap in Java. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing strong problem-solving skills in programming.

Additional Resources

For further reading and practice, consider the following resources: