Find missing number (1 to N)?

Short answer: int[] arr = { 1, 2, 4, 5 }; int n = 5; int expectedSum = n * (n + 1) / 2; int actualSum = 0; foreach (int num in arr) actualSum += num; int missing = expectedSum - actualSum;

Example code

int[] arr = { 1, 2, 4, 5 };
int n = 5;
int expectedSum = n * (n + 1) / 2;
int actualSum = 0;
foreach (int num in arr)
actualSum += num;
int missing = expectedSum - actualSum;

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.