Two Sum II - O(n log n) Time Complexity in Python

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 log n) time and use O(n) extra space.

Understanding the Problem

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

Potential pitfalls include assuming that there might be multiple solutions or using the same index twice, which the problem explicitly forbids.

Approach

To solve this problem, we can use a hash map to store the indices of the elements we have seen so far. This allows us to check in constant time whether the complement of the current element (i.e., target - current element) exists in the array.

Let's discuss a naive solution first:

Naive Solution

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

Optimized Solution

We can optimize the solution using a hash map to achieve O(n) time complexity. Here’s the thought process:

Initialize an empty hash map.
Iterate through the array, and for each element, calculate its complement (target - current element).
Check if the complement exists in the hash map. If it does, return the indices of the current element and its complement.
If the complement does not exist, add the current element and its index to the hash map.

Algorithm

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

Initialize an empty hash map called num_to_index.
Iterate through the array using a for loop.
For each element, calculate its complement as target - nums[i].
Check if the complement exists in num_to_index.
If it exists, return the indices [num_to_index[complement], i].
If it does not exist, add the element and its index to num_to_index.
If no solution is found by the end of the loop, return [-1, -1].

Code Implementation

def two_sum(nums, target):
    # Initialize an empty hash map
    num_to_index = {}
    
    # Iterate through the array
    for i, num in enumerate(nums):
        # Calculate the complement
        complement = target - num
        
        # Check if the complement exists in the hash map
        if complement in num_to_index:
            # Return the indices of the complement and the current number
            return [num_to_index[complement], i]
        
        # Add the current number and its index to the hash map
        num_to_index[num] = i
    
    # If no solution is found, return [-1, -1]
    return [-1, -1]

# Example usage
nums = [2, 7, 11, 15]
target = 9
print(two_sum(nums, target))  # Output: [0, 1]

Complexity Analysis

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

Edge Cases

Potential edge cases include:

An empty array: The function should return [-1, -1].
An array with one element: The function should return [-1, -1].
No two elements sum up to the target: The function should return [-1, -1].

Examples:

print(two_sum([], 9))  # Output: [-1, -1]
print(two_sum([1], 9))  # Output: [-1, -1]
print(two_sum([1, 2, 3], 7))  # Output: [-1, -1]

Testing

To test the solution comprehensively, consider a variety of test cases:

Simple cases with small arrays.
Cases where no solution exists.
Edge cases with empty or single-element arrays.
Cases with negative numbers and zero.

Example test cases:

print(two_sum([2, 7, 11, 15], 9))  # Output: [0, 1]
print(two_sum([3, 2, 4], 6))  # Output: [1, 2]
print(two_sum([3, 3], 6))  # Output: [0, 1]
print(two_sum([1, 2, 3], 7))  # Output: [-1, -1]

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Understand the problem requirements and constraints thoroughly.
Think about different approaches and their time and space complexities.
Use hash maps or other data structures to optimize your solution.
Practice solving similar problems to improve your problem-solving skills.

Conclusion

In this blog post, we discussed the Two Sum II problem, explored a naive solution, and then optimized it using a hash map. We also covered edge cases, testing, and provided tips for problem-solving. Understanding and solving such problems is crucial for improving your algorithmic thinking and coding skills.

Additional Resources

For further reading and practice, consider the following resources: