You are given a list of integers, nums, and an integer target. You need to find two distinct indices i and j such that the sum of the elements at those indices is equal to the target. You can assume that there will always be exactly one solution.

Write a function, `find_target_indices(nums: List[int], target: int) -> Tuple[int, int]`, that takes in the following parameters:
- `nums` (1 ≤ len(nums) ≤ 10^5): a list of integers, where each integer N (-10^9 ≤ N ≤ 10^9) represents an element in the array.
- `target` (-10^9 ≤ target ≤ 10^9): an integer representing the desired sum.

The function should return a tuple of two integers, (i, j), where i and j are the indices of the two elements that sum up to the target. The indices i and j should be 0-based.

Example:

Input:
nums = [2, 7, 11, 15]
target = 9

Output:
(0, 1)

Explanation:
The elements at indices 0 and 1 (2 and 7) sum up to the target value of 9.
user
Sure, here's a Python function that solves this problem using a HashTable to keep track of the prefix sums. The idea is simple - we calculate the prefix sums of `nums`, which are defined as `prefix_sum[i] = sum(nums[:i+1])`. Then, if a value is in the hash table, we can calculate its corresponding element quickly.

```python
class Solution:
    def find_target_indices(self, nums, target):
        prefix_sum = [0] + [prefix_sum[i-1] + nums[i-1] for I in range(1, len(nums) + 1)]
        seen = {}
        for I, num in enumerate(nums):
            Complement = target - num
            if Complement in seen:
                return i, seen[Complement]
            seen[num] = i
        return -1, -1
```

Here's how this function works:

1. It first creates a prefix sum list, which contains the sum of all the numbers starting from the beginning.
2. It then uses a hash table (in Python called a dictionary) to store the prefix sum and find it easier to join it.
3. If it is not an already owned number, it goes to the hash table, which produces the given result with the key of the moment.
4. If a key (Complement) is in the hash table, we calculate the necessary I, which shows their distance.
5. If a key isn't there, it means that no such pair exists, and the function returns -1, -1.

This method retrieves the answer in O(n) time complexity.
assistant
You are given a list of integers, nums, and an integer target. You need to find two distinct indices i and j such that the sum of the elements at those indices is equal to the target. You can assume that there will always be exactly one solution.

Write a function, `find_target_indices(nums: List[int], target: int) -> Tuple[int, int]`, that takes in the following parameters:
- `nums` (1 ≤ len(nums) ≤ 10^5): a list of integers, where each integer N (-10^9 ≤ N ≤ 10^9) represents an element in the array.
- `target` (-10^9 ≤ target ≤ 10^9): an integer representing the desired sum.

The function should return a tuple of two integers, (i, j), where i and j are the indices of the two elements that sum up to the target. The indices i and j should be 0-based.

Example:

Input:
nums = [2, 7, 11, 15]
target = 9

Output:
(0, 1)

Explanation:
The elements at indices 0 and 1 (2 and 7) sum up to the target value of 9.
user
Sure, I can help you write a Python function for this problem. This is a classic two-sum problem which can be efficiently solved using a hash map (dictionary in Python). The idea is to iterate through the list, and for each element, check if its complement (target - current element) exists in the dictionary. If it does, then we have found our two numbers. If it doesn't, we add the current element to the dictionary.

Here's the Python function:

```python
from typing import List, Tuple

def find_target_indices(nums: List[int], target: int) -> Tuple[int, int]:
    # Create an empty dictionary
    num_dict = {}

    # Iterate over the list
    for i, num in enumerate(nums):
        # Check if the complement exists in the dictionary
        complement = target - num
        if complement in num_dict:
            # If it does, return the current index and the index of the complement
            return (num_dict[complement], i)
        else:
            # If it doesn't, add the current number to the dictionary
            num_dict[num] = i

    # If no two numbers sum up to the target, raise an exception
    raise ValueError("No two numbers sum up to the target")
```

You can call this function with your example input like this:

```python
print(find_target_indices([2, 7, 11, 15], 9))
```

This will output `(0, 1)`, indicating that the numbers at indices 0 and 1 sum up to the target.
assistant