Compute square root to 1 decimal

Problem

Given a non-negative real number x, implement a function sqrt1dp(x) that returns $\sqrt{x}$ rounded (or truncated—clarify with interviewer) to one digit after the decimal point.

Requirements

• Do not call a built-in square root function.
• Your answer must be accurate to 1 decimal place (e.g., error < 0.05 if rounding).
• Discuss how you would optimize the algorithm (time complexity and convergence).

Examples

• x = 2 → 1.4
• x = 9 → 3.0
• x = 0 → 0.0

Clarifications to ask

• Rounding vs truncation to 1 decimal.
• Input range (e.g., up to 1e9?) and whether x can be non-integer.
• Acceptable error tolerance if not using decimal formatting.

Compute Square Root to 1 Decimal Place

Given a non-negative real number x, implement sqrt1dp(x) that returns the square root of x rounded to one digit after the decimal point, WITHOUT calling any built-in square-root function (no math.sqrt, Math.sqrt, **0.5, pow(x, 0.5), etc.).

The result must be accurate to 1 decimal place (rounding error < 0.05).

Approach: Use binary search (or Newton's method) to converge on the root, then round to one decimal. Binary search over the interval [0, max(1, x)]: each iteration halves the candidate window, so ~60 iterations already drives the error far below 1e-9. Newton's iteration g = (g + x/g) / 2 converges quadratically and is even faster.

Examples:

• sqrt1dp(2) → 1.4
• sqrt1dp(9) → 3.0
• sqrt1dp(0) → 0.0
• sqrt1dp(10) → 3.2

Constraints

• 0 <= x <= 1e9
• x may be a non-integer real number
• Do not call any built-in square-root function
• Result must be accurate to 1 decimal place (error < 0.05)

Hints

1. The root of x lies in [0, max(1, x)] — note that for 0 < x < 1 the root is GREATER than x, which is why the upper bound is max(1, x), not x.
2. Binary search: pick mid; if mid*mid < x the root is higher, else it is lower. Each iteration halves the error, so a fixed iteration count (e.g. 100-200) guarantees far better than 1-decimal precision.
3. Newton's method converges faster: start with a guess g and repeat g = (g + x/g) / 2 until it stops changing.
4. Only round to 1 decimal AFTER converging to full precision — rounding early loses accuracy. Use round(root, 1).