K-th Smallest Remaining Even Integer in Subarray Queries Python/Java solution

In Python or Java, the usual implementation builds a merge sort tree where each node stores a sorted list of even numbers from its segment. Queries traverse relevant nodes and perform binary searches inside their lists to count elements and locate the k-th smallest candidate.

What is the best approach for K-th Smallest Remaining Even Integer in Subarray Queries?

The most practical approach uses a merge sort tree or another range order-statistic structure. Each node stores sorted even values for its segment, allowing queries to count how many elements are ≤ a candidate value. Combined with binary search on the answer, each query runs in O(log² n) time while preprocessing takes O(n log n).

Is K-th Smallest Remaining Even Integer in Subarray Queries asked at Google/Amazon/Meta?

Variants of this problem appear in interviews at companies like Google, Amazon, and Meta under range query or order-statistic themes. Interviewers often frame it as 'k-th smallest number in a subarray' with extra constraints such as filtering by parity.

What data structure is used in K-th Smallest Remaining Even Integer in Subarray Queries?

Typical solutions rely on segment trees, merge sort trees, wavelet trees, or order-statistic balanced trees. These structures support efficient range queries and rank counting, which are required to locate the k-th smallest value among filtered elements.

What is the time complexity of K-th Smallest Remaining Even Integer in Subarray Queries?

Brute force scanning and sorting per query costs O(q * n log n). Optimized solutions using merge sort trees handle each query in O(log² n) after O(n log n) preprocessing. Wavelet trees or specialized order-statistic structures can reduce query time to about O(log n).

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3911. K-th Smallest Remaining Even Integer in Subarray Queries

Q: Is K-th Smallest Remaining Even Integer in Subarray Queries easy or hard?

The problem is typically categorized as Hard because it combines range queries with order statistics and filtering conditions. Efficient solutions require knowledge of segment trees, binary search on the answer, or advanced structures such as wavelet trees.

Q: How to solve K-th Smallest Remaining Even Integer in Subarray Queries in O(n)?

Pure O(n) per query is generally not feasible because the task requires both range filtering and order statistics. The closest practical optimization is preprocessing with advanced structures like wavelet trees, giving O(log n) query time after O(n log n) build time.

Hard28.6% acceptance

You are given an integer array nums where nums is strictly increasing.

You are also given a 2D integer array queries, where queries[i] = [l_i, r_i, k_i].

For each query [l_i, r_i, k_i]:

Consider the subarray nums[l_i..r_i]
From the infinite sequence of all positive even integers: 2, 4, 6, 8, 10, 12, 14, ...
Remove all elements that appear in the subarray nums[l_i..r_i].
Find the k_i^th smallest integer remaining in the sequence after the removals.

Return an integer array ans, where ans[i] is the result for the i^th query.

Example 1:

Input: nums = [1,4,7], queries = [[0,2,1],[1,1,2],[0,0,3]]

Output: [2,6,6]

Explanation:

`i`	`queries[i]`	`nums[l_i..r_i]`	Removed Evens	Remaining Evens	`k_i`	`ans[i]`
0	[0, 2, 1]	[1, 4, 7]	[4]	2, 6, 8, ...	1	2
1	[1, 1, 2]	[4]	[4]	2, 6, 8, ...	2	6
2	[0, 0, 3]	[1]	[]	2, 4, 6, ...	3	6

Thus, ans = [2, 6, 6].

Example 2:

Input: nums = [2,5,8], queries = [[0,1,2],[1,2,1],[0,2,4]]

Output: [6,2,12]

Explanation:

`i`	`queries[i]`	`nums[l_i..r_i]`	Removed Evens	Remaining Evens	`k_i`	`ans[i]`
0	[0, 1, 2]	[2, 5]	[2]	4, 6, 8, ...	2	6
1	[1, 2, 1]	[5, 8]	[8]	2, 4, 6, ...	1	2
2	[0, 2, 4]	[2, 5, 8]	[2, 8]	4, 6, 10, 12, ...	4	12

Thus, ans = [6, 2, 12].

Example 3:

Input: nums = [3,6], queries = [[0,1,1],[1,1,3]]

Output: [2,8]

Explanation:

`i`	`queries[i]`	`nums[l_i..r_i]`	Removed Evens	Remaining Evens	`k_i`	`ans[i]`
0	[0, 1, 1]	[3, 6]	[6]	2, 4, 8, ...	1	2
1	[1, 1, 3]	[6]	[6]	2, 4, 8, ...	3	8

Thus, ans = [2, 8].

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
nums is strictly increasing
1 <= queries.length <= 10⁵
queries[i] = [l_i, r_i, k_i]
0 <= l_i <= r_i < nums.length
1 <= k_i <= 10⁹

Topics

Array Binary Search

Hints open one at a time

Hint 1

Binary search on the answer

Hint 2

For each query, since nums is strictly increasing, you can find the count of even integers in the range

Hint 3

Subtract the count of even integers in the range from the current number you are checking in the binary search, and check whether, after removing the remaining numbers, this becomes the k-th

LanguageLoading editor…

1class Solution:
2    def kthRemainingInteger(self, nums: list[int], queries: list[list[int]]) -> list[int]:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[1,4,7]
[[0,2,1],[1,1,2],[0,0,3]]

Input

[2,5,8]
[[0,1,2],[1,2,1],[0,2,4]]

Input

[3,6]
[[0,1,1],[1,1,3]]

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