Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value in O(log N) time (Find First and Last Position).

Optimal Modified Binary Search Approach:
- Execute two targeted binary searches: one finding the leftmost boundary, and one finding the rightmost boundary.

def search_range(nums: list[int], target: int) -> list[int]:
    def find_bound(is_first: bool) -> int:
        left, right = 0, len(nums) - 1
        bound = -1
        while left <= right:
            mid = (left + right) // 2
            if nums[mid] == target:
                bound = mid
                if is_first:
                    right = mid - 1  # Keep searching left
                else:
                    left = mid + 1   # Keep searching right
            elif nums[mid] < target:
                left = mid + 1
            else:
                right = mid - 1
        return bound
    start = find_bound(True)
    if start == -1:
        return [-1, -1]
    end = find_bound(False)
    return [start, end]

Complexity & Key Logic:
- Time Complexity: O(log N) using two standard binary search passes.
- Space Complexity: O(1) constant auxiliary space.
- Algorithm Details:
- When nums[mid] == target, record mid and continue narrowing the search boundary (right = mid - 1 for first, left = mid + 1 for last).