Given that: int[] list = {4,6,12,18,21,27,32,39,42,47,53,59,64,67,73}; and: int targetValue = 24; and given this Binary Search algorithm: public static int BinarySearch(int[] list, int targetValue) { int mid, low, high; low = 0; high = list.length - 1; while (high >= low) { mid = (high + low) / 2; if (list[mid] < targetValue) { low = mid + 1; } else if (list[mid] > targetValue) { high = mid - 1; } else { return mid; } } return -1; // not found } Which value in list will be the third to be compared to the targetValue in a binary search for the target value?简答题
登录即可查看完整答案
我们收录了全球超50000道真实原题与详细解析,现在登录,立即获得答案。

Question

Given that:

int[] list = {4,6,12,18,21,27,32,39,42,47,53,59,64,67,73};

and:

int targetValue = 24;
and given this Binary Search algorithm:
public static int BinarySearch(int[] list, int targetValue) {
  int mid, low, high;

low = 0;
  high = list.length - 1;

while (high >= low) {
    mid = (high + low) / 2;
    if (list[mid] < targetValue) {
      low = mid + 1;
    } else if (list[mid] > targetValue) {
      high = mid - 1;
    } else {
      return mid;
    }
  }

return -1; // not found
}

Which value in list will be the third to be compared to the targetValue in a binary search for the target value?

BlackTom AI · Accepted Answer

Binary search compares the middle element of the current search interval (using integer division) and narrows the interval accordingly.

1. Relevant knowledge points and concepts
- The array is sorted and indexed from 0 to 14 (15 elements).
- mid is computed as (low + high) / 2 using integer division (floor).
- Each comparison inspects list[mid] and then updates low or high.
- We must track the sequence of mid indices and their values until the third comparison.

2. Step-by-step reasoning with intermediate calculations
Initial bounds: low = 0, high = 14.

First comparison:
- mid = (low + high) / 2 = (0 + 14) / 2 = 7.
- list[7] = 39.
- Compare 39 to targetValue 24: 39 > 24, so set high = mid - 1 = 6.
(First compared value: 39)

Second comparison:
- low = 0, high = 6.
- mid = (0 + 6) / 2 = 3.
- list[3] = 18.
- Compare 18 to 24: 18 < 24, so set low = mid + 1 = 4.
(Second compared value: 18)

Third comparison:
- low = 4, high = 6.
- mid = (4 + 6) / 2 = 5.
- list[5] = 27.
- Compare 27 to 24: 27 > 24, so set high = mid - 1 = 4.
(Third compared value: 27)

3. Calculations shown: (0+14)/2 = 7 → list[7]=39; (0+6)/2 = 3 → list[3]=18; (4+6)/2 = 5 → list[5]=27.

4. Why the correct answer is correct
The third inspected midpoint index is 5, and list[5] equals 27, so 27 is the third value compared during the binary search for 24.

5. Why other likely values are incorrect
- 39 was the first comparison, not the third.
- 18 was the second comparison.
- 21 or 32 might be plausible nearby values but they are at indices 4 and 6 and are not the third mid chosen (the mid chosen is index 5).

6. Conclusion
The third value compared during the binary search is 27, which matches the provided answer: 27.

类似问题

Binary Search Considering the following list. {12, 13, 15, 20, 23, 24, 25, 36, 40} What are the two first elements of the list that will be checked to find 25 using binary search? 1: The 1st element 2: The 2nd element

Given the following list of sorted elements, how many elements of the list will be checked to find 25 using binary search? {12, 13, 15, 20, 23, 24, 25, 36, 40}

What is the runtime complexity of the following code? NumberSearch(numbers, N, key) { mid = 0; low = 0; high = N - 1; while (high >= low) { mid = (high + low) / 2; if (numbers[mid] < key) low = mid + 1; else if (numbers[mid] > key) high = mid – 1; else return mid; } return -1; }

What is the prerequisite for using binary search on a list?

Binary search can be implemented using which programming approaches?

What happens if binary search is used on an unsorted list?

In a binary search algorithm, if the middle element is greater than the target, what happens next?

Given a list of 64 elements, how many elements will be checked to look for a value that is larger than the largest value in the list using binary search?

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单