Consider an array $A$ of integers of size $n$. The indices of $A$ run from $1$ to $n$. An algorithm is to be designed to check whether $A$ satisfies the condition given below. $$\forall i, j \in \{1, \dots, n - 1\} \text{ such that } i > j, (A[i + 1] - A[i]) > (A[j + 1] - A[j])$$ Which one of the following gives the worst case time complexity of the fastest algorithm that can be designed for the problem?

Let $D[k] = A[k + 1] - A[k]$ for all $k \in \{1, 2, \dots, n - 1\}$.

The given condition requires that for all $i > j$:

This is equivalent to stating that the sequence of adjacent differences $D$ is strictly increasing:

To check this property:

1. It is both necessary and sufficient to verify that $D[k] < D[k + 1]$ for every $k \in \{1, \dots, n - 2\}$, which is equivalent to $(A[k + 1] - A[k]) < (A[k + 2] - A[k + 1])$.

2. We can perform a single linear scan over the array $A$ from index $1$ to $n - 2$ and check the condition at each step.

3. This linear scan takes $\Theta(n)$ time in the worst case.

Since any algorithm must inspect $\Omega(n)$ elements of the array in the worst case to verify this condition, the optimal worst-case time complexity is $\Theta(n)$.

Therefore, (A) is the correct option.