Problem Description
You are given an integer array gifts
denoting the number of gifts in various piles. Every second, you do the following:
- Choose the pile with the maximum number of gifts.
- If there is more than one pile with the maximum number of gifts, choose any.
- Leave behind the floor of the square root of the number of gifts in the pile. Take the rest of the gifts.
Return the number of gifts remaining after k
seconds.
Example 1:
Input: gifts = [25,64,9,4,100], k = 4 Output: 29 Explanation: The gifts are taken in the following way:
- In the first second, the last pile is chosen and 10 gifts are left behind.
- Then the second pile is chosen and 8 gifts are left behind.
- After that the first pile is chosen and 5 gifts are left behind.
- Finally, the last pile is chosen again and 3 gifts are left behind. The final remaining gifts are [5,8,9,4,3], so the total number of gifts remaining is 29.
Example 2:
Input: gifts = [1,1,1,1], k = 4 Output: 4 Explanation: In this case, regardless which pile you choose, you have to leave behind 1 gift in each pile. That is, you can’t take any pile with you. So, the total gifts remaining are 4.
Constraints:
1 <= gifts.length <= 103
1 <= gifts[i] <= 109
1 <= k <= 103
Difficulty: Easy
Tags: array, heap (priority queue), simulation
Rating: 91.67%
Solution
Here’s my Python solution to this problem:
class Solution:
def pickGifts(self, gifts: List[int], k: int) -> int:
n = len(gifts)
gifts = [-g for g in gifts]
heapq.heapify(gifts)
for _ in range(k):
m = -heapq.heappop(gifts)
heapq.heappush(gifts, -math.floor(math.sqrt(m)))
return -sum(gifts)
Complexity Analysis
The solution has the following complexity characteristics:
- Time Complexity:
- Initial heapification:
- operations per operation: O(k log n)
- Total:
- Space Complexity:
- Modify the input array in place