LeetCode 2148: Count Elements With Strictly Smaller and Greater Elements Solution

Master LeetCode problem 2148 (Count Elements With Strictly Smaller and Greater Elements ), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2148. Count Elements With Strictly Smaller and Greater Elements

Easy

Array Sorting Counting

Problem Explanation

Explanation:

To solve this problem, we can iterate through the array and check for each element if it has a strictly smaller and strictly greater element. We can maintain two flags to keep track of whether we have found a smaller and a greater element for each element. We can then count the number of elements that satisfy this condition.

Sort the array to easily find the smaller and greater elements for each element.
Iterate through the sorted array and keep track of whether we have found a smaller and a greater element for each element.
Count the number of elements that satisfy the condition.

Time Complexity:

The time complexity of this approach is O(nlogn) due to sorting the array.

Space Complexity:

The space complexity is O(n) for storing the sorted array.

Solution Code

import java.util.Arrays;

class Solution {
    public int countElements(int[] nums) {
        int n = nums.length;
        int count = 0;
        
        Arrays.sort(nums);
        
        for (int i = 1; i < n - 1; i++) {
            if (nums[i] > nums[0] && nums[i] < nums[n - 1]) {
                count++;
            }
        }
        
        return count;
    }
}

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What is the time complexity of LeetCode 2148 (Count Elements With Strictly Smaller and Greater Elements )?

The time complexity for LeetCode 2148 (Count Elements With Strictly Smaller and Greater Elements ) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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