Asked in: SALESFORCE
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All Testcases Passed ✔
Solution
from collections import defaultdict
def getSpecialSubarrays(efficiency, k):
target_median = efficiency[k-1]
k -= 1
// ... rest of solution available after purchaseExplanation
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To approach this problem, you need to focus on a few key ideas: understanding how medians work in subarrays of odd lengths, and finding an efficient way to count how many such subarrays have a specific value as the median. Let’s explore the logical path step-by-step.
First, recognize that the problem revolves around medians of **odd-length subarrays** that must **include the element at index k** and have **efficiency[k] as the median**. The condition that efficiency[k] is unique in the array simplifies our job when identifying valid subarrays.
Let’s begin by recalling that in any sorted array of odd length `L`, the median is the `(L + 1)/2`-th smallest element. So, for a subarray to have efficiency[k] as its median, it must contain efficiency[k], and in the sorted version of that subarray, efficiency[k] must land at position `(L + 1)/2`. That means it must be greater than or equal to exactly `(L - 1)/2` elements in that subarray, and less than or equal to the same number of elements (since median is the middle value in a sorted list).
So, one way to think about the problem is to generate all possible odd-length subarrays that include the index `k`, and count how many of them have efficiency[k] at the median position. But given that the array can be quite large (up to 300,000 elements), we can’t afford to examine all possible subarrays directly. Therefore, we need to find a more analytical way to count them.
A good direction of thinking is to transform the array in a way that makes it easier to evaluate potential subarrays. Since we're comparing elements to efficiency[k], we can replace each value in the array with a transformed value:
- If an element is greater than efficiency[k], we can treat it as +1
- If it is equal to efficiency[k], treat it as 0
- If it is less than efficiency[k], treat it as -1
This transformation helps reduce the problem to a more manageable mathematical structure. Why? Because for any subarray, the median position in the sorted version is determined by the balance of elements that are smaller, larger, or equal to the median. By converting values to +1, 0, and -1, we can represent subarrays using prefix sums, and track how the balance of higher vs. lower elements affects the position of the median.
Now, because efficiency[k] is the required median and appears only once, the subarrays we care about must contain index `k`, and be of odd length, centered in some way around `k` so that efficiency[k] ends up as the middle value. So, now the challenge becomes identifying how many such subarrays exist that meet the required balance: equal numbers of +1s and -1s around the central 0.
This balance can be tracked using prefix sums from left and right of index k. For prefix sums, we can process the array to the left of k (moving backward), and to the right of k (moving forward), calculating running balances. You would use a hash map to track how many times each balance occurs on the left side. Then, as you process the right side, you can calculate how many combinations of left and right balances result in a total balance that puts efficiency[k] in the median position. This is essentially counting how many times the cumulative sum of transformed values is zero (or meets a specific condition) across subarrays that include k and are of odd length.
The odd-length requirement also introduces a parity condition. For example, if the subarray length is odd, the number of elements to the left and right of index k must be equal. So, you must consider combinations of left and right subarrays that, together with index k, result in an odd total length. This parity consideration limits which prefix sum pairs you match.
To summarize, your approach should focus on:
1. Understanding what makes efficiency[k] the median of a subarray: its position and the number of smaller and larger elements around it.
2. Transforming the array into a simplified representation based on its relationship to efficiency[k].
3. Using prefix sums to analyze the balance between elements greater and less than efficiency[k] around index k.
4. Efficiently counting how many such combinations of prefix sums from the left and right sides result in subarrays where efficiency[k] is the median and the length is odd.
Avoid brute force, as it will be too slow. Instead, use hashing, prefix sums, and balance counting to achieve an efficient solution. If you can implement the above ideas carefully, you’ll be able to solve the problem even for the largest input sizes.
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