Example: FindMissingNumberSummation in Java

// FindMissingNumberSummation
import java.util.Arrays; // Not strictly needed for this approach, but often imported

public class Main {
    public static void main(String[] args) {
        int[] arr1 = {3, 1, 5, 2}; // Missing 4
        System.out.println("Array: " + Arrays.toString(arr1) + ", Missing number: " + findMissingNumberSummation(arr1));

        int[] arr2 = {1, 2, 3, 5, 6}; // Missing 4
        System.out.println("Array: " + Arrays.toString(arr2) + ", Missing number: " + findMissingNumberSummation(arr2));

        int[] arr3 = {2, 3, 4, 5}; // Missing 1
        System.out.println("Array: " + Arrays.toString(arr3) + ", Missing number: " + findMissingNumberSummation(arr3));
    }

    /**
     * Finds the missing number in an array using the summation method.
     * Assumes numbers from 1 to n+1, with one number missing.
     *
     * @param arr The input array of distinct integers.
     * @return The missing integer.
     */
    public static int findMissingNumberSummation(int[] arr) {
        // Step 1: Determine 'n', which is the count of present numbers.
        // The complete sequence would have n + 1 numbers (from 1 to n+1).
        int n = arr.length;

        // Step 2: Calculate the expected sum of numbers from 1 to (n+1).
        // The formula for the sum of an arithmetic series 1 to K is K * (K + 1) / 2.
        // Here, K = n + 1.
        long expectedSum = (long) (n + 1) * (n + 2) / 2;

        // Step 3: Calculate the actual sum of elements present in the array.
        long actualSum = 0;
        for (int num : arr) {
            actualSum += num;
        }

        // Step 4: The difference is the missing number.
        return (int) (expectedSum - actualSum);
    }
}