You have an array of floating point numbers averages
which is initially empty. You are given an array nums
of n
integers where n
is even.
You repeat the following procedure n / 2
times:
minElement
, and the largest element maxElement
, from nums
.(minElement + maxElement) / 2
to averages
.Return the minimum element in averages
.
Example 1:
Input: nums = [7,8,3,4,15,13,4,1]
Output: 5.5
Explanation:
step | nums | averages |
---|---|---|
0 | [7,8,3,4,15,13,4,1] | [] |
1 | [7,8,3,4,13,4] | [8] |
2 | [7,8,4,4] | [8,8] |
3 | [7,4] | [8,8,6] |
4 | [] | [8,8,6,5.5] |
The smallest element of averages, 5.5, is returned.
Example 2:
Input: nums = [1,9,8,3,10,5]
Output: 5.5
Explanation:
step | nums | averages |
---|---|---|
0 | [1,9,8,3,10,5] | [] |
1 | [9,8,3,5] | [5.5] |
2 | [8,5] | [5.5,6] |
3 | [] | [5.5,6,6.5] |
Example 3:
Input: nums = [1,2,3,7,8,9]
Output: 5.0
Explanation:
step | nums | averages |
---|---|---|
0 | [1,2,3,7,8,9] | [] |
1 | [2,3,7,8] | [5] |
2 | [3,7] | [5,5] |
3 | [] | [5,5,5] |
2 <= n == nums.length <= 50
n
is even.1 <= nums[i] <= 50
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The brute force approach is like trying every single possible combination to find the best one. In this problem, we want to find a small section of numbers where the average of the smallest and largest numbers within that section is the smallest across all possible sections.
Here's how the algorithm would work step-by-step:
def minimum_average_of_smallest_and_largest_elements_brute_force(numbers):
minimum_average = float('inf')
# Iterate through all possible sub-arrays.
for start_index in range(len(numbers)):
for end_index in range(start_index, len(numbers)):
sub_array = numbers[start_index:end_index + 1]
# Find the smallest and largest numbers.
smallest_number = min(sub_array)
largest_number = max(sub_array)
# Calculate the average.
current_average = (smallest_number + largest_number) / 2
# Update the minimum average if necessary.
if current_average < minimum_average:
# Keep track of the minimum average
minimum_average = current_average
return minimum_average
The core idea is to cleverly sort and pair numbers to minimize the average. We focus on matching the smallest numbers with the largest numbers to keep the averages low from the start.
Here's how the algorithm would work step-by-step:
def find_minimum_average_of_smallest_and_largest_elements(number_list):
number_list.sort()
minimum_average = float('inf')
# Pair smallest with largest to minimize averages
for i in range(len(number_list) // 2):
current_average = (number_list[i] + number_list[len(number_list) - 1 - i]) / 2
# Compare against the lowest average found so far.
minimum_average = min(minimum_average, current_average)
if len(number_list) % 2 != 0:
# If odd number of elements, account for middle element.
middle_element = number_list[len(number_list) // 2]
minimum_average = min(minimum_average, middle_element)
return minimum_average
Case | How to Handle |
---|---|
Empty input array | Return an empty list or null to indicate no minimum average can be computed. |
Array with only one element | Return an empty list or null, as a minimum average requires at least two numbers. |
Array with all identical elements | The algorithm should correctly compute the average of any pair of these identical numbers, selecting the minimum. |
Array with only two elements | Calculate and return the average of these two elements as the only possible result. |
Array containing very large numbers leading to potential integer overflow | Use a data type capable of holding the sum of two large numbers (e.g., long, BigInteger), and divide using floating-point to avoid integer division issues. |
Array with a large number of elements (scalability) | Optimize the algorithm for time complexity (e.g., avoid nested loops if possible) and consider space complexity for auxiliary data structures. |
Array containing negative numbers | The algorithm should correctly handle negative numbers when calculating the minimum average, including comparing positive and negative averages. |
Array containing zero | Zero should be handled correctly when calculating the averages, including comparing averages with zero or negative averages. |