Computers and Technology, 09.04.2021 20:30 19thomasar
Consider the following method, which implements a recursive binary search.
/** Returns an index in nums where target appears if target
* appears in nums between nums[lo] and nums[hi], inclusive;
* otherwise, returns -1.
* Precondition: nums is sorted in ascending order.
* lo >= 0, hi < nums. length, nums. length > 0
*/
public static int bSearch(int[] nums, int lo, int hi, int target)
{
if (hi >= lo)
{
int mid = (lo + hi) / 2;
if (nums[mid] == target)
{
return mid;
}
if (nums[mid] > target)
{
return bSearch(nums, lo, mid - 1, target);
}
else
{
return bSearch(nums, mid + 1, hi, target);
}
}
return -1;
}
The following code segment appears in a method in the same class as bSearch.
int target = 3;
int[] nums = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20};
int targetIndex = bSearch(nums, 0, nums. length - 1, target);
How many times will bSearch be called as a result of executing the code segment above?
1- A
2- B
3- C
4-D
5-E
Answers: 2
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Consider the following method, which implements a recursive binary search.
/** Returns an index in...
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