Robert BaindourovNodeJS, ReactJS, ElectronJS - Web Software Architect

Summary: To find the final array state after k multiplication operations, you can use a min-heap (priority queue). By storing elements as (value, index) pairs, you can efficiently extract the smallest element, multiply it by the given multiplier, and reinsert it in O(log n) time per operation, resulting in an optimal O(n log n + k log n) total time complexity.

You are given an integer array nums, an integer k, and an integer multiplier.

You need to perform k operations on nums. In each operation:

• Find the minimum value x in nums. If there are multiple occurrences of the minimum value, select the one that appears first.
• Replace the selected minimum value x with x * multiplier.

Return an integer array denoting the final state of nums after performing all k operations.

Approach Using a Min-Heap

We use a priority queue (min-heap) that stores pairs (value, index) to efficiently find the smallest number in the array along with its index.

Steps:

1. Push all elements of nums into the min-heap. Each element is stored as a tuple (value, index) (to preserve the index).
2. Perform k operations:
   • Extract the smallest element (top of the heap) from the min-heap.
   • Update the value at the extracted index in nums by multiplying it with multiplier.
   • Push the updated value (new_value, index) back into the heap.
3. After all k operations, return the modified nums array.

Priority Queue Configuration:

• Since we want to retrieve the smallest element, we configure priority_queue to behave as a min-heap using greater<tuple<int, int>>.

#include <iostream>
#include <vector>
#include <queue>
#include <tuple>
using namespace std;

vector<int> getFinalState(vector<int>& nums, int k, int multiplier) {
    // Min-heap to store pairs of (value, index)
    priority_queue<tuple<int, int>, vector<tuple<int, int>>, greater<tuple<int, int>>> minHeap;

    // Step 1: Push all elements into the min-heap
    for (int i = 0; i < nums.size(); i++) {
        minHeap.push({nums[i], i});
    }

    // Step 2: Perform k operations
    for (int i = 0; i < k; i++) {
        // Extract the smallest element from the min-heap
        auto [value, index] = minHeap.top(); // Get the top element (smallest)
        minHeap.pop(); // Remove it from the heap

        // Update the value in the original array
        nums[index] = value * multiplier;

        // Push the updated value back into the min-heap
        minHeap.push({nums[index], index});
    }

    // Step 3: Return the modified array
    return nums;
}

int main() {
    vector<int> nums = {2, 3, 1, 5, 1}; // Example array
    int k = 2;                          // Number of operations
    int multiplier = 4;                 // Multiplier

    vector<int> result = getFinalState(nums, k, multiplier);

    // Output the result
    for (int num : result) {
        cout << num << " ";
    }

    return 0;
}

Explanation of the Code:

1. Heap Initialization:
   • All elements in nums are pushed into the min-heap along with their respective indices. This ensures that when we extract the minimum element, we also know its index in the original array.
2. Perform k Operations:
   • For each operation, the smallest value from the heap is fetched (using minHeap.top()).
   • This value is updated (multiplied by multiplier) in the nums array.
   • The updated value, along with the same index, is reinserted into the heap to maintain the order of elements.
3. Final State:
   • After k operations, the array nums is modified, and we return it.

Complexity Analysis:

1. Time Complexity:
   • Building the initial heap: O(n log n) (inserting n elements into the heap).
   • For each of the k operations:
     • Extract the minimum: O(log n)
     • Reinsert the updated value: O(log n).
   • Total: O(n log n + k log n).
   • In most cases, if k is small relative to n, the runtime is dominated by O(n log n).
2. Space Complexity:
   • The priority queue (min-heap) uses O(n) space to store all the elements.

Frequently Asked Questions (FAQ)

Why use a min-heap instead of sorting the array each time?

Sorting the array repeatedly for k operations would result in a time complexity of O(k * n log n), which is highly inefficient for large arrays. Using a min-heap (priority queue) optimizes the process to O(n log n + k log n) by keeping the elements partially sorted and allowing O(log n) minimum extraction and insertion.

Can we use a max-heap for this problem?

No, a max-heap would place the largest elements at the top. Since the problem explicitly requires us to find and modify the minimum value in each operation, a min-heap is the correct data structure.

What happens if there are multiple identical minimum values?

If there are multiple occurrences of the minimum value, the problem statement requires selecting the one that appears first in the original array. By storing tuples of (value, index) in our min-heap, we ensure that if values are equal, the smaller index is given priority, satisfying this requirement automatically.