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Tested Python solution for LeetCode 3264 with 18 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 3264 # by problem number
lcpy gen -s final_array_state_after_k_multiplication_operations_i # by problem name
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.nums after performing all k operations.
Input: nums = [2,1,3,5,6], k = 5, multiplier = 2
Output: [8,4,6,5,6]
| 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] |
Input: nums = [1,2], k = 3, multiplier = 4
Output: [16,8]
| Operation | Result |
|---|---|
| After operation 1 | [4, 2] |
| After operation 2 | [4, 8] |
| After operation 3 | [16, 8] |
import heapq
class Solution:
# Time: O(n + k * log n)
# Space: O(n)
def get_final_state(self, nums: list[int], k: int, multiplier: int) -> list[int]:
heap = [(value, index) for index, value in enumerate(nums)]
heapq.heapify(heap)
for _ in range(k):
value, index = heapq.heappop(heap)
heapq.heappush(heap, (value * multiplier, index))
result = [0] * len(nums)
while heap:
value, index = heapq.heappop(heap)
result[index] = value
return result
| Time | Space |
|---|---|
| O(n + k * log n) | O(n) |