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

The problem is generally classified as Medium difficulty. The core idea—tracking the k-th largest element—requires understanding heap behavior and streaming updates. Developers familiar with priority queues can implement the optimal solution quickly.

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

Python solutions typically use the heapq module to maintain a min-heap, while Java implementations use PriorityQueue. In both languages, the algorithm inserts each new value, removes the smallest element if the heap exceeds size k, and reads the heap root to determine the k-th largest element.

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

Maintain a min-heap containing at most k elements. Insert every new value into the heap and remove the smallest element if the heap grows beyond size k. After each insertion, the heap's root represents the current k-th largest element, which can be used to update the power value immediately.

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

The most efficient approach uses a min-heap (priority queue) that stores only the largest k elements seen so far. Each insertion pushes a value into the heap and removes the smallest if the size exceeds k. The heap root always represents the k-th largest element, allowing constant-time access for power updates. This results in O(n log k) time and O(k) space.

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

Variants of k-th largest element streaming problems appear frequently in interviews at companies like Amazon, Google, and Meta. Interviewers often test the ability to maintain running statistics using heaps or priority queues. The same technique appears in problems such as 'Kth Largest Element in a Stream'.

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

A min-heap (priority queue) is the primary data structure used to track the k largest elements efficiently. The heap keeps the smallest among those k elements at the root, which corresponds to the current k-th largest value in the full dataset.

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

Using the optimal min-heap approach, each insertion requires O(log k) time because heap operations are logarithmic in the heap size. Processing n insertions therefore takes O(n log k). The space complexity is O(k) since the heap stores only the k largest elements.

3935. Power Update After K-th Largest Insertion I - 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] and the difference between consecutive k_i values is always less than 10.

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] <= 10⁶
1 <= p <= 10⁶
1 <= queries.length <= 2 × 10⁴
1 <= val_i <= 10⁶
1 <= k_i <= n + i + 1
|k_i - k_{i - 1}| < 10 for i > 0