How many comparisons does binary search make in the worst case for an array of s

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Question: How many comparisons does binary search make in the worst case for an array of size 16?

Options:

Correct Answer: 7

Solution:

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

How many comparisons does binary search make in the worst case for an array of size 16?

How many comparisons does binary search make in the worst case 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 made is 4 + 1 = 5.

Binary Search – A search algorithm that finds the position of a target value within a sorted array by repeatedly dividing the search interval in half.
Logarithmic Complexity – The time complexity of binary search is O(log n), which indicates that the number of comparisons grows logarithmically with the size of the array.
Worst-Case Scenario – The maximum number of comparisons needed to find an element or determine its absence in the array.