To solve the problem of counting elements that have both a strictly smaller and a strictly greater element in the list, we can follow these steps:
Here’s an implementation of the described strategy:
def count_elements(nums):
if len(nums) < 3:
# If there are less than 3 elements, there can't be an element with both
# a strictly smaller and a strictly greater element.
return 0
# Sort the list to easily get the minimum and maximum values.
sorted_nums = sorted(nums)
min_val = sorted_nums[0]
max_val = sorted_nums[-1]
count = 0
for num in nums:
if num > min_val and num < max_val:
count += 1
return count
# Example usage:
nums = [11, 7, 2, 15]
print(count_elements(nums)) # Output: 2
nums = [-3, 3, 3, 90, -3]
print(count_elements(nums)) # Output: 1
Thus, the overall time complexity of the solution is (O(n \log n)).
If there are no further clarifications needed, this should suffice as an efficient and comprehensive solution to the problem.
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