Find the Missing Number in an Array (1..n)

📌 Definition:

📖 Detailed Explanation:

Use the arithmetic series: the expected sum of 1..n is n(n+1)/2; subtract the actual array sum and the difference is the missing number. Use a wide type to avoid overflow, or XOR all of 1..n with all elements so pairs cancel and the missing value remains.

💻 Solutions:

• Java
• JavaScript
• Python

public static int missing(int[] a, int n) {
    long expected = (long) n * (n + 1) / 2, actual = 0;
    for (int x : a) actual += x;
    return (int) (expected - actual);
}

function missing(a, n) {
  const expected = n * (n + 1) / 2;
  return expected - a.reduce((s, x) => s + x, 0);
}

def missing(a, n):
    return n * (n + 1) // 2 - sum(a)

🔑 Key Points:

• Expected sum n(n+1)/2 minus actual sum
• Use a wide type to avoid overflow
• XOR variant is overflow-proof
• O(n) time, O(1) space

🌍 Real-World Example: