Question

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.
∀𝑖, 𝑗∈{1, … , 𝑛−1} 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?

• A. Θ(𝑛)
• B. Θ(log(𝑛))
• C. Θ(𝑛 log(𝑛))
• D. \(\Theta(n^2)\)

Hint:

The condition says the difference sequence \(d[k]=A[k+1]-A[k]\) is strictly increasing. This only requires checking consecutive pairs, doable in one linear scan, giving \(\Theta(n)\).

Answer 1

The Correct Option is A

Step 1: Rewrite the condition using consecutive differences. Let \(d[k] = A[k+1]-A[k]\) for \(k=1\) to \(n-1\). The problem asks us to verify that whenever one index exceeds another, its difference value also exceeds the other's difference value -- in other words, the array of differences must be sorted in strictly ascending order.
Step 2: A naive approach would test every pair \((i,j)\) with \(i > j\), giving \(O(n^2)\) comparisons, but this ignores the transitive nature of the 'less than' ordering.
Step 3: Because 'ascending order' is a transitive property, testing only adjacent pairs \(d[k] < d[k+1]\) for \(k = 1, \dots, n-2\) is both necessary and sufficient. If any adjacent pair fails, the sequence is not ascending and the original condition fails for some \((i,j)\); if all adjacent pairs pass, ascending order (and hence the original condition) is guaranteed for every pair by chaining inequalities.
Step 4: Both computing the \(n-1\) differences and checking the \(n-2\) adjacent comparisons take linear time, so the whole check runs in \(O(n)\).
Step 5: Since reading every element of the input array is unavoidable to guarantee correctness in the worst case (an adversary could hide a violation at any single position), \(\Omega(n)\) is also a lower bound.
Step 6: The matching upper and lower bounds pin the worst case complexity of the optimal algorithm at \(\Theta(n)\).
Final Answer: \(\Theta(n)\)