Solving 'Final Array State After K Multiplication Operations' : Leetcode Solution

In this article, we will discuss a solution to the LeetCode problem "Final Array State After K Multiplication Operations". This problem challenges us to repeatedly multiply the smallest element in an array by a given multiplier and return the final state of the array after K operations.

Problem Description

You are given an array nums, an integer k, and a multiplier. Your task is to:

1. Perform K operations where in each operation you:

   • Select the smallest element in nums.

   • Multiply it by multiplier.

   • Update the element in the array.

2. Return the final state of the array after all operations.

Approach: Leveraging a Min-Heap

The main challenge in this problem is efficiently finding and updating the smallest element in the array K times. A min-heap is the perfect data structure for this purpose since it allows:

• O(1) access to the smallest element.

• O(log n) time for insertion and deletion.

Here are the steps to solve the problem:

1. Initialize a Min-Heap:

   • Use the heapq module in Python to transform the array into a min-heap. Instead of directly storing the values, we store tuples (value, index) to track the original positions of elements.

2. Iterate K Times:

   • Extract the smallest element from the heap using heapq.heappop.

   • Multiply it by the given multiplier.

   • Update the corresponding value in the result array.

   • Push the updated value back into the heap.

3. Return the Result:

   • After K iterations, return the updated array.

Python Code Implementation

Here’s the complete Python implementation:

import heapq

class Solution(object):
    def getFinalState(self, nums, k, multiplier):
        # Create a copy of nums to hold the results
        result = nums[::]

        # Build a min-heap with (value, index) pairs
        min_heap = [(n, i) for i, n in enumerate(nums)]
        heapq.heapify(min_heap)

        # Perform K operations
        for _ in range(k):
            # Extract the smallest value and its index
            n, i = heapq.heappop(min_heap)

            # Update the value in the result array
            result[i] *= multiplier

            # Push the updated value back into the heap
            heapq.heappush(min_heap, (result[i], i))

        return result

Example Walk-through

Let’s walk through an example to better understand the solution.

Input:

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

Execution:

1. Initial Min-Heap:

    [(1, 1), (2, 0), (3, 2), (5, 3), (6, 4)]
   

2. First Operation:

   • Pop (1, 1).

   • Multiply by 2 -> result[1] = 2.

   • Push (2, 1) back into the heap.

3. Second Operation:

   • Pop (2, 0).

   • Multiply by 2 -> result[0] = 4.

   • Push (4, 0) back into the heap.

4. Repeat until K operations are performed.

Final Result:

[4, 16, 12, 10, 6]

Complexity Analysis

• Time Complexity:

  • Building the min-heap: O(n).

  • K operations of pop and push: O(k log n).

  • Total: O(n + k log n).

• Space Complexity:

  • Min-heap storage: O(n).

Conclusion

This approach efficiently handles the requirement to repeatedly update the smallest element in the array while keeping track of indices. By leveraging the heapq module, we ensure that the solution is both time-efficient and concise. The flexibility of Python’s heap implementation makes this problem straightforward once you understand the heap operations.