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Binary search on a float range was the move, but I kept second-guessing my termination condition.
Use Newton's method (or binary search) to iteratively approximate the square root, stopping when the change is less than 10^-precise. Then analyze the time complexity based on the number of iterations and the cost of arithmetic operations.
Pro tip: Mention that Newton's method converges quadratically, so the number of iterations is logarithmic in the required precision, making it efficient for high precision. Also, clarify how you handle edge cases like val=0 and precise=0.
Confirm that val and precise are non-negative integers, and that the result must be a floating-point number with absolute error ≤ 10^-precise. Discuss handling val=0 and precise=0.
Select Newton's method for quadratic convergence or binary search for simplicity. Explain the trade-offs: Newton's method is faster but requires careful initialization; binary search is robust but slower.
For Newton's method, start with an initial guess (e.g., val or 1) and iterate x = (x + val/x)/2 until the change is less than 10^-precise. For binary search, set low=0, high=max(1, val), and narrow the interval until the width is less than 10^-precise.
For Newton's method, the number of iterations is O(log(precise)) due to quadratic convergence, and each iteration involves constant-time arithmetic operations (assuming fixed-precision arithmetic). For binary search, the number of iterations is O(precise) because the interval halves each time, leading to O(precise) iterations.
Address floating-point precision issues, such as when val is very large or precise is high. Mention using epsilon = 10^-precise for termination and ensuring the error bound is met.
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