My first instinct was to just iterate from 1 upward and check each square, which obviously fails the time constraint.
Use binary search on the range [1, n] to find an integer whose square equals n. At each step, compare mid*mid with n and adjust the search bounds accordingly. This yields O(log n) time and avoids built-in square root functions.
Pro tip: Mention that you can optimize the search space to [1, n/2] for n > 1, and use long long to prevent integer overflow when computing mid*mid. This shows attention to edge cases and performance.
Confirm that n is a positive integer and discuss handling of n=1 (perfect square) and n=0 if allowed. Also address potential integer overflow in mid*mid.
Set low=1 and high=n (or n/2 for n>1) to cover all possible integer square roots. Explain why these bounds are valid.
While low <= high, compute mid, then compare mid*mid with n. If equal, return true; if less, set low=mid+1; else set high=mid-1.
If the loop ends without finding an exact square, return false. Discuss why the algorithm correctly determines non-squares.
State that time complexity is O(log n) due to halving the search space, and space is O(1). Suggest testing with perfect squares, non-squares, and large values.
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