Best-case time complexity of binary search on a sorted array of length n?

• A. O(1)
• B. O(log n)
• C. O(n)
• D. O(n log n)

Answer: (B)

The best-case scenario for binary search is constant time because you can find the target in the very first comparison. In a sorted array, binary search checks the middle element first; if that element is the target, you’re done after just one comparison, regardless of n. Therefore the time growth does not depend on the array size, giving O(1). In other cases, you typically halve the search space each step, leading to about log2(n) comparisons (O(log n)) on average or in the worst case, and not every situation will involve linear or higher work like O(n) or O(n log n).