Is Count Smaller Elements With Opposite Parity easy or hard?

The problem is usually categorized as Medium because it combines two ideas: ordering elements and filtering by parity. The brute force solution is simple, but recognizing that sorting or prefix counting reduces the complexity is the key step.

Count Smaller Elements With Opposite Parity Python/Java solution

Typical implementations sort the array and maintain two counters: one for even numbers and one for odd numbers. While scanning the sorted list, query the counter of the opposite parity to compute the result, then update the current parity counter. This pattern translates directly to Python, Java, and C++ with identical logic.

How to solve Count Smaller Elements With Opposite Parity in O(n)?

Pure O(n) solutions are uncommon unless the value range is small enough for counting arrays. If values are bounded, you can maintain frequency counts for even and odd numbers and compute prefix totals directly. In the general case with arbitrary integers, sorting or a Fenwick Tree leads to O(n log n) complexity.

What is the best approach for Count Smaller Elements With Opposite Parity?

The most practical solution sorts the numbers and keeps running counts of odd and even values encountered so far. Because the array is processed in ascending order, every processed element is guaranteed to be smaller. Checking parity determines which counter to use. This approach runs in O(n log n) time due to sorting and uses O(n) space.

Is Count Smaller Elements With Opposite Parity asked at Google/Amazon/Meta?

Problems involving counting smaller elements and parity filtering appear frequently in technical interviews. Variants of this pattern show up at companies like Amazon, Google, and Meta when discussing order statistics, prefix counting, or Binary Indexed Tree usage.

What data structure is used in Count Smaller Elements With Opposite Parity?

Common implementations rely on sorting plus simple counters for odd and even values. More advanced solutions use a Fenwick Tree (Binary Indexed Tree) or segment tree to support efficient prefix queries while tracking parity groups.

What is the time complexity of Count Smaller Elements With Opposite Parity?

The brute force method compares every pair of elements and runs in O(n^2) time with O(1) space. The optimized solution sorts the array and tracks parity counts, reducing the complexity to O(n log n). Advanced implementations using Fenwick Trees also run in O(n log n).

3907. Count Smaller Elements With Opposite Parity - Practice Online

You are given an integer array nums of length n.

The score of an index i is defined as the number of indices j such that:

i < j < n
nums[j] < nums[i]
nums[i] and nums[j] have different parity (one is even and the other is odd).

Return an integer array answer of length n, where answer[i] is the score of index i.

Example 1:

Input: nums = [5,2,4,1,3]

Output: [2,1,2,0,0]

Explanation:

For i = 0, the elements nums[1] = 2 and nums[2] = 4 are smaller and have different parity.
For i = 1, the element nums[3] = 1 is smaller and has different parity.
For i = 2, the elements nums[3] = 1 and nums[4] = 3 are smaller and have different parity.
No valid elements exist for the remaining indices.

Thus, the answer = [2, 1, 2, 0, 0].

Example 2:

Input: nums = [4,4,1]

Output: [1,1,0]

Explanation:

For i = 0 and i = 1, the element nums[2] = 1 is smaller and has different parity. Thus, the answer = [1, 1, 0].

Example 3:

Input: nums = [7]

Output: [0]

Explanation:

No elements exist to the right of index 0, so its score is 0. Thus, the answer = [0].

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹