What is the worst-case number of comparisons in binary search for an array of si

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Question: What is the worst-case number of comparisons in binary search for an array of size 16?

Options:

Correct Answer: 6

Solution:

The worst-case number of comparisons is log2(16) = 4, but since we start counting from 0, it takes 5 comparisons.

What is the worst-case number of comparisons in binary search for an array of size 16?

What is the worst-case number of comparisons in binary search for an array of size 16?

Step 1: Understand what binary search is. It is a method to find an item in a sorted array by repeatedly dividing the search interval in half.
Step 2: Know the size of the array. In this case, the array size is 16.
Step 3: Use the formula for the worst-case number of comparisons in binary search, which is log2(n), where n is the size of the array.
Step 4: Calculate log2(16). Since 16 is 2 raised to the power of 4 (2^4 = 16), log2(16) equals 4.
Step 5: Remember that we start counting comparisons from 0. So, if log2(16) is 4, the total number of comparisons needed is 4 + 1 = 5.
Step 6: Conclude that the worst-case number of comparisons in binary search for an array of size 16 is 5.

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