The binary search angle is pretty clear once you think about it for a second, but the part that actually tripped me up was the overflow.
Use binary search on the range [1, n] to find an integer x such that x^2 equals n. At each step, compute mid^2 and compare with n, adjusting the search range accordingly. This yields O(log n) time and avoids built-in square root functions.
Pro tip: Mention that you can optimize by setting the upper bound to n/2 for n > 1, and use integer division to avoid overflow when computing mid. Also, clarify that you're using binary search on the answer space, not on the input array.
Confirm that n is a positive integer and discuss edge cases like n=1 (perfect square) and n=2 (not). Also consider potential integer overflow when squaring mid.
Set low=1 and high=n. Optionally, optimize high to n/2 for n>1 to reduce search space, but ensure correctness for n=1.
While low <= high, compute mid = low + (high - low)/2. Compare mid*mid with n: if equal, return true; if less, set low = mid+1; if greater, set high = mid-1.
If the loop ends without finding a perfect square, return false. Ensure the algorithm runs in O(log n) time.
Explain that the search space halves each iteration, giving O(log n) time and O(1) space. Walk through examples like n=16 (true) and n=14 (false).
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