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69. Sqrt(x)

easyAsked at Goldman Sachs

Compute the integer part of sqrt(x) without using a built-in sqrt function. Goldman Sachs uses this to test whether you can implement binary search on the answer space — the fundamental quant interview skill.

By Sam K., Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Goldman Sachs loops.

  • Glassdoor (2026-Q1)— Goldman Sachs Strats and Engineering reports list Sqrt(x) as a 'binary search on the answer space' question.
  • LeetCode Discuss (2025-10)— Sqrt(x) sits in the top-20 of LeetCode's Goldman Sachs company tag.

Problem

Given a non-negative integer x, return the square root of x rounded down to the nearest integer. The returned integer should be non-negative as well. You must not use any built-in exponent function or operator. For example, do not use pow(x, 0.5) in C++ or x ** 0.5 in Python.

Constraints

  • 0 <= x <= 2^31 - 1

Examples

Example 1

Input
x = 4
Output
2

Example 2

Input
x = 8
Output
2

Explanation: sqrt(8) is 2.82..., rounded down is 2.

Approaches

1. Linear scan

Try k = 0, 1, 2, ... until k*k > x; return k-1.

Time
O(sqrt(x))
Space
O(1)
function mySqrtLinear(x) {
  let k = 0;
  while ((k + 1) * (k + 1) <= x) k++;
  return k;
}

Tradeoff: Correct but O(sqrt(x)) — at x = 2^31 - 1 it does ~46000 iterations, which times out on Goldman's HackerRank timer. Mention briefly then move to binary search.

2. Binary search on the answer (optimal)

The answer is in [0, x]. Binary-search for the largest k such that k*k <= x.

Time
O(log x)
Space
O(1)
function mySqrt(x) {
  if (x < 2) return x;
  let lo = 1, hi = Math.floor(x / 2);
  while (lo <= hi) {
    const mid = Math.floor(lo + (hi - lo) / 2);
    const sq = mid * mid;
    if (sq === x) return mid;
    if (sq < x) lo = mid + 1;
    else hi = mid - 1;
  }
  return hi;
}

Tradeoff: O(log x) — about 31 iterations at INT_MAX. The 'return hi' at the end is the part Goldman wants to see — after the loop, hi is the floor of the true square root.

3. Newton's method

Iterate x_{n+1} = (x_n + x / x_n) / 2 until convergence.

Time
O(log x)
Space
O(1)
function mySqrtNewton(x) {
  if (x < 2) return x;
  let r = x;
  while (r * r > x) {
    r = Math.floor((r + Math.floor(x / r)) / 2);
  }
  return r;
}

Tradeoff: Converges quadratically — fewer iterations than binary search in practice. Mention as a bonus to show numerical-analysis chops; binary search is what most Goldman interviewers expect.

Goldman Sachs-specific tips

Goldman Sachs uses Sqrt(x) specifically to test 'binary search on the answer space' as a concept. The pivot from 'I need sqrt' to 'I'll binary-search for the largest k with k*k <= x' is the moment that earns the credit. The lo + (hi - lo) / 2 instead of (lo + hi) / 2 detail is the small thing Goldman looks for — overflow safety in 32-bit languages.

Common mistakes

  • Overflow on mid*mid for large x — in JS this is fine due to 64-bit floats, but Goldman's Java/C++ rounds will require BigInteger or the (long) cast.
  • Returning lo instead of hi after the loop — at exit, hi < lo and hi is the floor of sqrt.
  • Forgetting the x < 2 short-circuit, which infinite-loops on x = 0 if lo and hi start at 1 and 0.

Follow-up questions

An interviewer at Goldman Sachs may pivot to one of these next:

  • Compute sqrt(x) to k decimal places. (Continue binary search past the integer boundary on a scaled value.)
  • Compute cube root of x. (Same template — binary search the answer.)
  • Newton-Raphson convergence rate vs binary search — when is Newton better?

Solve it now

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Output

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FAQ

Why does Goldman ban Math.sqrt or pow?

Because the question is about your ability to do binary search on the answer space — not your knowledge of the standard library. The 'no exponent function' rule is in the problem statement specifically to force this.

Why mid * mid <= x and not (mid + 1) * (mid + 1) <= x?

Both work; they're different invariants. The version above keeps the invariant 'lo is a candidate answer that might be too small, hi is a candidate that might be too large' and reads cleanly. Either is correct.

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