Can the sum formula overflow an int?

Yes, for large n. n * (n + 1) passes Integer.MAX_VALUE once n reaches 46,341, and Java wraps silently instead of throwing. This problem stops at n = 5,000, but doing the math in long (as above) or using XOR removes the risk entirely.

How do I avoid overflow without switching to long?

Fold the two sums into one running difference with no division: int missing = nums.length; then missing += i - nums[i]; for every index. Java int arithmetic is defined to wrap modulo 2³², so even if the running value wraps on a huge input, the final result is exact because the true answer fits in an int. The formula version cannot use that trick: n * (n + 1) wraps before the / 2, and the division does not undo the wrap. XOR sidesteps the question entirely.

Why copy the array before sorting it?

Arrays.sort sorts in place, so the caller would see their array reordered. Arrays.copyOf(nums, nums.length) or nums.clone() gives you a private copy for O(n) extra memory.

Is a HashSet a good alternative?

It works: add every value to a HashSet , then return the first i from 0 to n that contains rejects. It is O(n) on average but boxes every element, so it is slower and heavier than the sum or XOR versions.

Can I write the sum with streams?

Yes: return nums.length * (nums.length + 1) / 2 - Arrays.stream(nums).sum(); is one line. IntStream.sum returns an int and wraps on overflow like the loop; use Arrays.stream(nums).asLongStream().sum() if you need a long.

Missing Number in an Array in Java: Sort, Sum and XOR Solutions

missing_number.java

Java 25.0

Not run yet

+ 6 hidden

Input

nums = [5, 2, 0, 3, 1]

Expected

Five numbers, so the range is 0 to 5. Every value shows up except 4.