You are tasked with implementing a function called `binary_search`, which performs a binary search on a sorted list of integers and returns the index of the target value if found, or -1 if the target value is not present in the list.

The function should have the following signature:
```python
def binary_search(nums: List[int], target: int) -> int:
    pass
```

**Input:**
- The input consists of two parameters:
  - `nums`: a list of integers (1 <= len(nums) <= 10^5) in ascending order, where the elements of the list are unique.
  - `target`: an integer value to search for in the list.

**Output:**
- The function should return an integer representing the index of the target value in the list, or -1 if the target is not present.

**Note:**
- You are not allowed to use the built-in function `list.index()`.
- You should implement the function using an iterative approach and not recursion.

**Examples:**
```
binary_search([1, 3, 5, 7, 9], 3) => 1
binary_search([1, 3, 5, 7, 9], 2) => -1
binary_search([1, 3, 5, 7, 9], 9) => 4
```
user
Sure, I'd be happy to help you with implementing the binary search function. Here's a Python implementation using the iterative approach:

```python
def binary_search(nums, target):
    left, right = 0, len(nums) - 1
    while left <= right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        elif nums[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1
```

In this implementation, the `left` and `right` variables are used to define the search range. The loop iterates until `left` is greater than `right`. During each iteration, the midpoint of the search range is calculated, and the target value is compared with the midpoint value. If the midpoint is equal to the target, its index is returned. If the midpoint is less than the target, the loop only searches in the right half of the search range. Similarly, if the midpoint is greater than the target, the loop only searches in the left half of the search range. Once the target is found or eliminated, `-1` is returned if the target is not found.

This algorithm's time complexity is O(logn), which is efficient and ideal for large input arrays. The space complexity required by this algorithm is O(1), which is favorable since sorting is not required in this algorithm.

This function does not use any recursion in the application, including list.index().
assistant
You are tasked with implementing a function called `binary_search`, which performs a binary search on a sorted list of integers and returns the index of the target value if found, or -1 if the target value is not present in the list.

The function should have the following signature:
```python
def binary_search(nums: List[int], target: int) -> int:
    pass
```

**Input:**
- The input consists of two parameters:
  - `nums`: a list of integers (1 <= len(nums) <= 10^5) in ascending order, where the elements of the list are unique.
  - `target`: an integer value to search for in the list.

**Output:**
- The function should return an integer representing the index of the target value in the list, or -1 if the target is not present.

**Note:**
- You are not allowed to use the built-in function `list.index()`.
- You should implement the function using an iterative approach and not recursion.

**Examples:**
```
binary_search([1, 3, 5, 7, 9], 3) => 1
binary_search([1, 3, 5, 7, 9], 2) => -1
binary_search([1, 3, 5, 7, 9], 9) => 4
```
user
Sure, here is the implementation of the `binary_search` function in Python. This function uses an iterative approach to perform binary search on a sorted list of integers.

```python
def binary_search(nums: List[int], target: int) -> int:
    left, right = 0, len(nums) - 1

    while left <= right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        elif nums[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1
```

In this function, `left` and `right` are two pointers that keep track of the search range in the list. The `mid` pointer is calculated as the average of `left` and `right`, and it represents the middle element of the current search range. If the target value is equal to the middle element, we return the index of the middle element. If the target value is greater than the middle element, we update `left` to be one position to the right of the middle element. If the target value is smaller than the middle element, we update `right` to be one position to the left of the middle element. We continue this process until we find the target value or the search range becomes empty, in which case we return -1.
assistant