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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:
x in nums. If there are multiple occurrences of the minimum value, select the one that appears first.x with x * multiplier.Return an integer array denoting the final state of nums after performing all k operations.
Example 1:
Input: nums = [2,1,3,5,6], k = 5, multiplier = 2
Output: [8,4,6,5,6]
Explanation:
| Operation | Result |
|---|---|
| After operation 1 | [2, 2, 3, 5, 6] |
| After operation 2 | [4, 2, 3, 5, 6] |
| After operation 3 | [4, 4, 3, 5, 6] |
| After operation 4 | [4, 4, 6, 5, 6] |
| After operation 5 | [8, 4, 6, 5, 6] |
Example 2:
Input: nums = [1,2], k = 3, multiplier = 4
Output: [16,8]
Explanation:
| Operation | Result |
|---|---|
| After operation 1 | [4, 2] |
| After operation 2 | [4, 8] |
| After operation 3 | [16, 8] |
Constraints:
1 <= nums.length <= 1001 <= nums[i] <= 1001 <= k <= 101 <= multiplier <= 5Problem Overview: You receive an integer array nums, an integer k, and a multiplier. Perform exactly k operations. In each operation, find the smallest value in the array (choose the smallest index if there is a tie), multiply it by multiplier, and update the array. After all operations, return the final state of the array.
This is mainly a simulation problem built around repeatedly selecting the minimum element. The main design decision is how efficiently you retrieve that minimum value during each operation.
Approach 1: Simple Iteration Approach (O(n * k) time, O(1) space)
The most straightforward solution simulates the process exactly as described. For each of the k operations, iterate through the entire array to find the smallest value and its index. Once found, multiply that element by multiplier and continue to the next operation. The algorithm uses a simple linear scan of the array each time the minimum must be identified.
This approach works well when k and n are small. The implementation is extremely simple and mirrors the problem statement directly. However, every operation requires scanning the entire array, resulting in O(n) work per operation and O(n * k) total time.
Approach 2: Using Min-Heap (Priority Queue) (O((n + k) log n) time, O(n) space)
A more efficient solution maintains the current minimum using a min-heap (priority queue). Push all elements into the heap as pairs (value, index). The heap automatically keeps the smallest value at the top, and the index ensures ties are resolved correctly.
During each operation, pop the smallest element from the heap, multiply its value by multiplier, update the corresponding index in the array, and push the updated pair back into the heap. Each pop and push operation costs O(log n), so performing k operations results in O(k log n) heap work after the initial O(n) heap construction.
This method avoids repeated full scans of the array and is ideal when k is large. It models the process as a simulation but relies on a heap to keep minimum lookups efficient.
Recommended for interviews: Start by explaining the direct simulation using iteration to demonstrate understanding of the rules. Then move to the min‑heap optimization. Interviewers typically expect the heap solution because it reduces repeated minimum searches and scales better when the number of operations grows.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Simple Iteration (Linear Scan) | O(n * k) | O(1) | Small arrays or small k where simplicity matters more than optimization |
| Min-Heap (Priority Queue) | O((n + k) log n) | O(n) | Large k or repeated minimum lookups where heap structure avoids full scans |