How to Quickly Find a Missing ID in a Sequence – Simple Sum and XOR Solutions

When a senior engineer from a large tech firm joins a smaller company, they often criticize the low technical level and chaotic processes, highlighting the cultural gap between big‑company systematic practices and small‑company survival tactics.

One frequent interview question asks candidates to identify a missing ID in a sequence that should be continuous, such as student IDs from 1 to 100 where one number is absent.

Interview Question: Find the Missing ID

The straightforward approach is to iterate through the array, but that becomes inefficient for millions of entries. A more efficient method uses the arithmetic series sum: compute the expected total = n × (n + 1) / 2, subtract the actual sum of the array, and the difference is the missing ID.

public class MissingIdFinder {
    public static int findMissing(int[] ids, int n) {
        int total = n * (n + 1) / 2;
        int sum = 0;
        for (int id : ids) {
            sum += id;
        }
        return total - sum;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 4, 5, 6}; // 3 is missing
        int missing = findMissing(arr, 6);
        System.out.println("Missing ID is: " + missing);
    }
}

Key points: n must represent the maximum expected ID, the array should contain no duplicates or out‑of‑range values, and for very large data sets consider using long to avoid integer overflow.

Another common solution employs the XOR operation, which also runs in O(n) time and O(1) space:

public static int findMissingXor(int[] ids, int n) {
    int xorAll = 0;
    for (int i = 1; i <= n; i++) {
        xorAll ^= i;
    }
    for (int id : ids) {
        xorAll ^= id;
    }
    return xorAll;
}

Both methods are simple, maintainable, and interview‑friendly. When discussing them, mention the time complexity O(n), space complexity O(1), and potential overflow concerns for large n. The choice between sum and XOR often depends on personal preference, but the sum approach is usually easier to explain.