Check if a number is a perfect square?

Short answer: public bool IsPerfectSquare(int num) { if (num < 0) return false; int left = 0, right = num; while (left <= right) { int mid = left + (right - left) / 2; long sq = (long)mid * mid; if (sq == num) return true; else if (sq < num) left = mid + 1; else right = mid - 1; } return false; } Explanation: Binary search for integer square root and check if square equals num.

Example code

public bool IsPerfectSquare(int num) {
if (num < 0) return false;
int left = 0, right = num; while (left <= right) { int mid = left + (right - left) / 2;
long sq = (long)mid * mid;
if (sq == num) return true;
else if (sq < num) left = mid + 1;
else right = mid - 1;
}
return false;
} Explanation: Binary search for integer square root and check if square equals num.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.