My first instinct was to just loop from 0 upward and check when i*i exceeds x.
Use binary search to find the largest integer whose square is less than or equal to x, handling edge cases like x=0 and x=1. Avoid overflow by comparing mid to x/mid instead of mid*mid, and clearly explain the time and space complexity.
Pro tip: Mention that you can use Newton's method for a faster convergence, but binary search is simpler and more reliable for interviews. Also, explicitly handle integer overflow to show attention to detail.
Confirm that x is non-negative and discuss edge cases: x=0 returns 0, x=1 returns 1. Also consider large values that might cause overflow.
Explain that binary search on the range [0, x] is efficient because the square root function is monotonic. Set low=0, high=x, and iterate while low <= high.
Compute mid = low + (high - low) // 2. To avoid overflow, compare mid with x // mid instead of computing mid*mid. Adjust low or high based on the comparison.
When the loop ends, high will be the integer square root rounded down. Return high.
State that time complexity is O(log x) and space complexity is O(1). Optionally, mention alternative approaches like Newton's method.
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