The correct recurrence for the worst case of Binary Search is T(n) = T(n/2) + O(1) and T(1) = T(0) = O(1).

Here's why:

1. In each step of a binary search, the algorithm divides the input into two equal halves. This is represented by T(n/2).

2. The algorithm performs a constant amount of work in each step (i.e., comparing the target value with the middle element), which is represented by O(1).

3. The base case is when the size of the input is 1 or 0, in which case the algorithm performs a constant amount of work, represented by T(1) = T(0) = O(1).

Question

The correct recurrence for the worst case of Binary Search is T(n) = T(n/2) + O(1) and T(1) = T(0) = O(1).

Here's why:

1. In each step of a binary search, the algorithm divides the input into two equal halves. This is represented by T(n/2).

2. The algorithm performs a constant amount of work in each step (i.e., comparing the target value with the middle element), which is represented by O(1).

3. The base case is when the size of the input is 1 or 0, in which case the algorithm performs a constant amount of work, represented by T(1) = T(0) = O(1).

Knowee AI · Accepted Answer

The correct recurrence for the worst case of Binary Search is T(n) = T(n/2) + O(1) and T(1) = T(0) = O(1).

Here's why:

1. In each step of a binary search, the algorithm divides the input into two equal halves. This is represented by T(n/2).

2. The algorithm performs a constant amount of work in each step (i.e., comparing the target value with the middle element), which is represented by O(1).

3. The base case is when the size of the input is 1 or 0, in which case the algorithm performs a constant amount of work, represented by T(1) = T(0) = O(1).

Which of the following is correct recurrence for worst case of Binary Search?T(n) = 2T(n/2) + O(1) and T(1) = T(0) = O(1)T(n) = T(n-1) + O(1) and T(1) = T(0) = O(1)T(n) = T(n/2) + O(1) and T(1) = T(0) = O(1)T(n) = T(n-2) + O(1) and T(1) = T(0) = O(1)

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