Find the missing number in a sequence from 1 to n using XOR?

Short answer: public int FindMissingNumber(int[] nums, int n) { int xor = 0; for (int i = 1; i <= n; i++) { xor ^= i; } foreach (int num in nums) { xor ^= num; } return xor; } Follow on: Explanation: XOR all numbers from 1 to n and XOR all elements in array; duplicates cancel out, leaving missing number.

Example code

public int FindMissingNumber(int[] nums, int n) {
int xor = 0;
for (int i = 1; i <= n; i++) {
xor ^= i;
}
foreach (int num in nums) {
xor ^= num;
}
return xor;
} Follow on: Explanation: XOR all numbers from 1 to n and XOR all elements in array; duplicates cancel out, leaving missing number.

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.