Is Power Update After K-th Largest Insertion II easy or hard?

Power Update After K-th Largest Insertion II is classified as a Hard problem because it combines streaming updates with order-statistic queries. Efficient solutions require familiarity with heaps, Fenwick trees, or segment trees and careful complexity management.

Power Update After K-th Largest Insertion II Python/Java solution

In Python, the solution typically uses the heapq module to maintain a min heap of size k. In Java, PriorityQueue provides the same behavior. For advanced implementations, Fenwick trees or segment trees are used to maintain dynamic order statistics.

How to solve Power Update After K-th Largest Insertion II in O(n log n)?

Use coordinate compression and maintain frequencies with a Fenwick Tree. Each insertion updates the tree in O(log n), and a binary search over prefix sums finds the k-th largest value. Additional prefix-sum queries compute any required power updates efficiently.

What is the best approach for Power Update After K-th Largest Insertion II?

The most practical solution maintains the k largest elements using a min heap of size k. Each insertion pushes a value into the heap and removes the smallest if the heap exceeds size k, keeping the root equal to the current k-th largest element. This keeps updates at O(log k) time while using O(k) space.

Is Power Update After K-th Largest Insertion II asked at Google/Amazon/Meta?

Problems involving maintaining the k-th largest element in a dynamic stream appear frequently in interviews at companies such as Google, Amazon, and Meta. Variations often combine heaps with additional data structures like segment trees or prefix sums to track extra metrics.

What data structure is used in Power Update After K-th Largest Insertion II?

Common choices include a min heap for tracking the top k elements and Fenwick Trees or segment trees for maintaining frequency counts and prefix sums. These structures support efficient rank queries and updates required after each insertion.

What is the time complexity of Power Update After K-th Largest Insertion II?

A brute force solution that sorts after every insertion costs O(n log n) per update. The optimized heap approach reduces this to O(log k) per insertion. If the power calculation requires aggregated statistics, Fenwick tree or segment tree solutions typically run in O(log n) per update.

3930. Power Update After K-th Largest Insertion II - Practice Online

You are given an integer array nums and an integer p.

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

For each query:

Insert val_i into nums.
Let x be the k_i^th largest element in the current nums.
Update p to p^x % (10⁹ + 7).

Return an array ans where the ans[i] represents the value of p after processing the i^th query.

Example 1:

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

Output: [64,4096]

Explanation:

`i`	`val_i`	Current `nums`	`k_i`	`k_i^th` largest	p	New `p = p^k % (10⁹ + 7)`
0	3	[2, 3]	1	3	4	4³ % (10⁹ + 7) = 64
1	1	[2, 3, 1]	2	2	64	64² % (10⁹ + 7) = 4096

Thus, ans = [64, 4096].

Example 2:

Input: nums = [7,5], p = 6, queries = [[4,3],[7,2]]

Output: [1296,220296870]

Explanation:

`i`	`val_i`	Current `nums`	`k_i`	`k_i^th` largest	`p`	New `p = p^k % (10⁹ + 7)`
0	4	[7, 5, 4]	3	4	6	6⁴ % (10⁹ + 7) = 1296
1	7	[7, 5, 4, 7]	2	7	1296	1296⁷ % (10⁹ + 7) = 220296870

Thus, ans = [1296, 220296870]

Constraints:

1 <= nums.length <= 2 * 10⁴
1 <= nums[i] <= 10⁹
1 <= p <= 10⁹
1 <= queries.length <= 2 * 10⁴
1 <= val_i <= 10⁹
1 <= k_i <= n + i + 1