#3264

Easy

LeetCode Problem Workspace

Final Array State After K Multiplication Operations I

Solve the problem of determining the final state of an array after multiple multiplication operations using a priority queue.

array math heap priority queue simulation

Problem Statement

You are given an integer array nums, an integer k, and an integer multiplier. The goal is to perform k operations on nums. In each operation, multiply one element by the multiplier and adjust the array's order accordingly. After completing k operations, return the final state of the array.

The operations should be performed sequentially, and the array must be updated each time by multiplying one of the elements by the multiplier. The task is to determine the state of the array after all k operations, considering the sorting behavior involved in these operations.

Examples

Example 1

Input: nums = [2,1,3,5,6], k = 5, multiplier = 2

Output: [8,4,6,5,6]

Example 2

Input: nums = [1,2], k = 3, multiplier = 4

Output: [16,8]

Constraints

1 <= nums.length <= 100
1 <= nums[i] <= 100
1 <= k <= 10
1 <= multiplier <= 5

Solution Approach

Using a Priority Queue

Store the elements of the array along with their indices as pairs in a priority queue. This will allow efficient access to the element that should be multiplied in each operation while maintaining the sorting of the array. After each multiplication, the array's state can be updated quickly.

Efficient Multiplication and Sorting

In each operation, apply the multiplier to one element, and then reinsert it into the priority queue. This maintains the correct order for the next multiplication step. The time complexity of each operation is logarithmic in terms of the array size due to the priority queue operations.

Simulating Operations

Perform exactly k operations, each involving selecting and multiplying one element in the array. The updated element is then placed back into the priority queue to ensure it is properly positioned for the next operation. This allows the simulation of all operations efficiently.

Complexity Analysis

Metric	Value
Time	O(N + k \cdot \log N)
Space	O(N)

The time complexity of the solution is O(N + k * log N), where N is the length of the array and k is the number of operations. The O(N) comes from initially building the priority queue, and the k * log N term arises from each operation requiring a log N time insertion and removal from the queue. The space complexity is O(N) due to the storage of the array and the priority queue.

What Interviewers Usually Probe

Look for the candidate's ability to handle the array transformations efficiently with a priority queue.
Check if the candidate understands the trade-offs of using a priority queue and its logarithmic time complexity.
Evaluate the candidate's approach to maintaining the array's sorted order throughout multiple operations.

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the solution by trying to manually sort the array after each operation, which increases time complexity.
Failing to account for the fact that the array size can vary up to 100 elements, making inefficient solutions impractical.
Not correctly updating the priority queue after each operation, leading to incorrect final results.

Follow-up variants

Increasing the number of operations k beyond the usual limits and testing if the approach still handles it efficiently.
Changing the multiplier value dynamically during the operations instead of keeping it constant.
Testing with different initial array sizes and ensuring the priority queue implementation scales well.

FAQ

What is the best approach for solving the Final Array State After K Multiplication Operations I problem?

Using a priority queue to maintain the order of elements is the most efficient approach. It ensures that the array is updated correctly after each operation, while maintaining a time complexity of O(N + k * log N).

How does the priority queue help in solving the Final Array State After K Multiplication Operations I problem?

The priority queue allows efficient access and modification of the element that needs to be multiplied, while maintaining the array's order in logarithmic time, which is crucial for handling multiple operations efficiently.

Can I solve this problem without using a priority queue?

While it's possible to solve the problem without a priority queue, doing so would likely increase the time complexity due to the need for sorting or searching through the array after each operation.

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

The time complexity is O(N + k * log N), where N is the length of the array and k is the number of operations. The priority queue ensures that each operation is handled in logarithmic time.

What should I focus on when solving the Final Array State After K Multiplication Operations I problem?

Focus on efficiently managing the array's order after each multiplication by using a priority queue. Ensure that the time complexity remains optimal by maintaining the array's sorted state during the operations.

terminal

Solution

Solution 1: Priority Queue (Min-Heap) + Simulation

We can use a min-heap to maintain the elements in the array $\textit{nums}$. Each time, we extract the minimum value from the min-heap, multiply it by $\textit{multiplier}$, and then put it back into the min-heap. During the implementation, we insert the indices of the elements into the min-heap and define a custom comparator function to sort the min-heap based on the values of the elements in $\textit{nums}$ as the primary key and the indices as the secondary key.

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class Solution:
    def getFinalState(self, nums: List[int], k: int, multiplier: int) -> List[int]:
        pq = [(x, i) for i, x in enumerate(nums)]
        heapify(pq)
        for _ in range(k):
            _, i = heappop(pq)
            nums[i] *= multiplier
            heappush(pq, (nums[i], i))
        return nums

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