My first instinct was to just iterate from 1 upward and check k*k == n, which works but is obviously O(sqrt n).
Use binary search to find the integer square root of n in the range [0, n]. At each step, compute mid*mid and compare with n, adjusting the search range accordingly. If an exact match is found, n is a perfect square; otherwise, it is not.
Pro tip: Mention that binary search is optimal for O(log n) and that using integer arithmetic avoids floating-point precision issues. Also, handle edge cases like n=0 and n=1 explicitly.
Confirm that n is non-negative and discuss edge cases such as n=0 and n=1, which are perfect squares. Also, consider the maximum value of n to ensure integer overflow is handled (e.g., use long in Java).
Explain that binary search on the range [0, n] yields O(log n) time. Alternatively, use Newton's method for integer square root, but binary search is simpler and meets the requirement.
Initialize low=0, high=n. While low <= high, compute mid = low + (high - low)/2. Compute square = mid*mid (using long to avoid overflow). If square == n, return true. If square < n, set low = mid + 1; else set high = mid - 1.
If the loop ends without finding an exact square, return false. State that time complexity is O(log n) and space complexity is O(1).
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