To solve the problem of finding the missing element in an array A consisting of N different integers from the range [1..(N + 1)], you can utilize the mathematical approach of calculating the expected sum versus the actual sum of the array elements.

Here’s the step-by-step explanation of the approach:

The sum of the first N+1 integers (which are [1, 2, ..., N+1]) can be calculated using the formula:

This formula gives the sum of all integers from 1 to N+1.

Compute the sum of all elements in array A.

The missing element will be the difference between the expected sum and the actual sum of array A.

Given the constraints (N can be up to 100,000), this approach is efficient with a time complexity of O(N) because it involves a single pass through the array to compute the sum.

Here is the implementation of the solution in JavaScript:

function solution(A) {
    const N = A.length;
    const expectedSum = (N + 1) * (N + 2) / 2;
    let actualSum = 0;
    
    for (let i = 0; i < N; i++) {
        actualSum += A[i];
    }
    
    const missingElement = expectedSum - actualSum;
    return missingElement;
}

// Example usage:
const A = [2, 3, 1, 5];
console.log(solution(A)); // Output: 4

Explanation of the Example:

For the array A = [2, 3, 1, 5], where N = 4 (since A.length = 4), the expected sum of [1, 2, 3, 4, 5] is 15.

The actual sum of elements in A is 2 + 3 + 1 + 5 = 11.

Therefore, the missing element is 15 - 11 = 4.

This approach ensures that you find the missing element in linear time, making it optimal for large inputs as specified in the problem constraints.