Two Number Sum

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice. You can return the answer in any order.

Break Down

1. Identify the input and output for the problem. In this case, the input is an array of integers and a target integer, and the output is the indices of the two numbers that add up to the target.
2. Determine which algorithm or data structure to use. There are several approaches to solve this problem, including the brute force approach, the hash table approach, and the sorting approach.
3. Write the code for your chosen approach, testing it thoroughly as you go along.
4. Run your solution on the test cases you identified in step 2, and verify that the output matches your expected results.
5. Optimize your solution if possible, by considering the time and space complexity of your code, and finding ways to reduce both if necessary.
6. Submit your solution and test it against other test cases, if available.

Remember to always test your solution thoroughly and consider edge cases to ensure that it works correctly in all scenarios.

Brute force solution:

Brute force solution: The simplest approach to the problem is to use a nested loop to iterate over all possible pairs of elements in the array and check if their sum is equal to the target. The time complexity of this approach is O(n^2), where n is the length of the array.

def two_sum_brute_force(nums, target):
    n = len(nums)
    for i in range(n):
        for j in range(i+1, n):
            if nums[i] + nums[j] == target:
                return [i, j]

Advantages:

• Easy to understand and implement
• Works for any kind of input array

Disadvantages:

• Slow for large arrays
• Not scalable for more complex problems

Hash table solution:

This solution involves using a hash table to store the indices of the elements in the array as keys, and their values as the indices. Then, we can iterate through the array and check if the complement of each element with respect to the target is already in the hash table. The time complexity of this approach is O(n), where n is the length of the array.

def two_sum_hash_table(nums, target):
    hash_table = {}
    for i, num in enumerate(nums):
        complement = target - num
        if complement in hash_table:
            return [hash_table[complement], i]
        hash_table[num] = i

Advantages:

• Faster than the brute force solution for large arrays
• Works for any kind of input array

Disadvantages:

• Uses extra memory to store the hash table
• May not work if there are duplicates in the input array

Sorting solution:

The Sorting solution sorts the array and uses two pointers to find the pair of elements that add up to the target sum. This solution has a time complexity of O(n log n) due to the sorting step.

def two_sum(nums, target):
    nums_sorted = sorted(nums)
    left = 0
    right = len(nums) - 1
    while left < right:
        if nums_sorted[left] + nums_sorted[right] == target:
            return [nums.index(nums_sorted[left]), nums.index(nums_sorted[right])]
        elif nums_sorted[left] + nums_sorted[right] < target:
            left += 1
        else:
            right -= 1
    return []

Advantages:

• More efficient than the brute force solution.
• Doesn’t require additional space like the hash map solution.

Disadvantages:

• Sorting can be expensive for large arrays.
• Not as simple to understand as the brute force solution.