What is the time complexity of Count Elements With Strictly Smaller and Greater Elements ?

The optimal solution for Count Elements With Strictly Smaller and Greater Elements runs in O(n) time and uses O(1) space. The time complexity is O(n) because we traverse the array a constant number of times (once for min/max and once for counting).

What is the brute force solution for Count Elements With Strictly Smaller and Greater Elements ?

The brute force approach for Count Elements With Strictly Smaller and Greater Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks each element in the array to see if there are both smaller and greater elements. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Elements With Strictly Smaller and Greater Elements ?

Count Elements With Strictly Smaller and Greater Elements uses the following concepts: Array, Sorting, Counting. The recommended patterns to study are: Array, Counting.

What are the interview tips for Count Elements With Strictly Smaller and Greater Elements ?

Always clarify the problem statement and constraints. Discuss your thought process before jumping into coding. Consider edge cases, such as arrays with all identical elements.

What are common mistakes when solving Count Elements With Strictly Smaller and Greater Elements ?

Not considering duplicates correctly when counting. Confusing the conditions for strictly smaller and greater.

#2148

Count Elements With Strictly Smaller and Greater Elements

Easy

ArraySortingCountingArrayCounting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

Instead of checking each element against all others, we can find the minimum and maximum values in the array. All elements except these two will have both smaller and greater elements.

⚙️

Algorithm

3 steps

1Step 1: Find the minimum and maximum values in the array.
2Step 2: Count how many times the minimum and maximum appear in the array.
3Step 3: Calculate the result as total elements minus the counts of min and max.

solution.py11 lines

1# Full working Python code
2
3def count_elements(nums):
4    min_val = min(nums)
5    max_val = max(nums)
6    count_min = nums.count(min_val)
7    count_max = nums.count(max_val)
8    return len(nums) - count_min - count_max
9
10# Example usage
11print(count_elements([11, 7, 2, 15]))  # Output: 2

ℹ

Complexity note: The time complexity is O(n) because we traverse the array a constant number of times (once for min/max and once for counting).

1Elements must be strictly smaller and greater, so we exclude min and max.
2Counting occurrences of min and max helps in calculating the result efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.