What is the time complexity of Final Array State After K Multiplication Operations II?

The optimal solution for Final Array State After K Multiplication Operations II runs in O(n + k log n) time and uses O(n) space. The time complexity is O(n + k log n) because we first build the heap in O(n) time, and each of the k operations takes O(log n) time for insertion and extraction.

What is the brute force solution for Final Array State After K Multiplication Operations II?

The brute force approach for Final Array State After K Multiplication Operations II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves repeatedly finding the minimum value in the array and multiplying it by the given multiplier for k times. This is straightforward but inefficient, especially for large k. The optimal Optimal Solution approach improves this to O(n + k log n).

What algorithm or data structure is used to solve Final Array State After K Multiplication Operations II?

Final Array State After K Multiplication Operations II uses the following concepts: Array, Heap (Priority Queue), Simulation. The recommended patterns to study are: Heap, Array.

What are the interview tips for Final Array State After K Multiplication Operations II?

Always analyze the time complexity of your approach. Discuss potential edge cases, such as large k or small array sizes. Explain your thought process clearly while coding.

What are common mistakes when solving Final Array State After K Multiplication Operations II?

Not considering the efficiency of finding the minimum value. Failing to apply modulo correctly after operations.

#3266

Final Array State After K Multiplication Operations II

Hard

ArrayHeap (Priority Queue)SimulationHeapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + k log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n + k log n)Space O(n)

The optimal solution leverages a priority queue (min-heap) to efficiently manage the minimum values in the array. This allows us to perform the k operations in logarithmic time, significantly reducing the overall complexity.

⚙️

Algorithm

3 steps

1Step 1: Initialize a min-heap with the elements of nums.
2Step 2: For k operations, extract the minimum element from the heap, multiply it by the multiplier, and push it back into the heap.
3Step 3: After k operations, extract all elements from the heap and apply modulo 10^9 + 7.

solution.py9 lines

1import heapq
2
3def finalArrayState(nums, k, multiplier):
4    MOD = 10**9 + 7
5    heapq.heapify(nums)
6    for _ in range(k):
7        min_val = heapq.heappop(nums)
8        heapq.heappush(nums, min_val * multiplier)
9    return [num % MOD for num in nums]

ℹ

Complexity note: The time complexity is O(n + k log n) because we first build the heap in O(n) time, and each of the k operations takes O(log n) time for insertion and extraction.

1Using a heap allows for efficient minimum retrieval and updates.
2Understanding the impact of large k values on performance is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.