Standard Priority Queue
A problem requiring a standard priority queue solution which can be solved in nlogn. Code is down below, cheers, ACC.
Maximal Score After Applying K Operations - LeetCode
2530. Maximal Score After Applying K Operations
Medium
You are given a 0-indexed integer array nums
and an integer k
. You have a starting score of 0
.
In one operation:
- choose an index
i
such that0 <= i < nums.length
, - increase your score by
nums[i]
, and - replace
nums[i]
withceil(nums[i] / 3)
.
Return the maximum possible score you can attain after applying exactly k
operations.
The ceiling function ceil(val)
is the least integer greater than or equal to val
.
Example 1:
Input: nums = [10,10,10,10,10], k = 5 Output: 50 Explanation: Apply the operation to each array element exactly once. The final score is 10 + 10 + 10 + 10 + 10 = 50.
Example 2:
Input: nums = [1,10,3,3,3], k = 3 Output: 17 Explanation: You can do the following operations: Operation 1: Select i = 1, so nums becomes [1,4,3,3,3]. Your score increases by 10. Operation 2: Select i = 1, so nums becomes [1,2,3,3,3]. Your score increases by 4. Operation 3: Select i = 2, so nums becomes [1,1,1,3,3]. Your score increases by 3. The final score is 10 + 4 + 3 = 17.
Constraints:
1 <= nums.length, k <= 105
1 <= nums[i] <= 109
public class Solution { public long MaxKelements(int[] nums, int k) { PriorityQueue pQueue = new PriorityQueue(false); long retVal = 0; foreach (int n in nums) pQueue.Enqueue((long)n, n); for (int i = 0; i < k; i++) { double temp = 0; long val = (long)pQueue.Dequeue(out temp); retVal += val; pQueue.Enqueue((long)Math.Ceiling(val / 3.0), (long)Math.Ceiling(val / 3.0)); } return retVal; } public class PriorityQueue { public struct HeapEntry { private object item; private double priority; public HeapEntry(object item, double priority) { this.item = item; this.priority = priority; } public object Item { get { return item; } } public double Priority { get { return priority; } } } private bool ascend; private int count; private int capacity; private HeapEntry[] heap; public int Count { get { return this.count; } } public PriorityQueue(bool ascend, int cap = -1) { capacity = 10000000; if (cap > 0) capacity = cap; heap = new HeapEntry[capacity]; this.ascend = ascend; } public object Dequeue(out double priority) { priority = heap[0].Priority; object result = heap[0].Item; count--; trickleDown(0, heap[count]); return result; } public object Peak(out double priority) { priority = heap[0].Priority; object result = heap[0].Item; return result; } public object Peak(/*out double priority*/) { //priority = heap[0].Priority; object result = heap[0].Item; return result; } public void Enqueue(object item, double priority) { count++; bubbleUp(count - 1, new HeapEntry(item, priority)); } private void bubbleUp(int index, HeapEntry he) { int parent = (index - 1) / 2; // note: (index > 0) means there is a parent if (this.ascend) { while ((index > 0) && (heap[parent].Priority > he.Priority)) { heap[index] = heap[parent]; index = parent; parent = (index - 1) / 2; } heap[index] = he; } else { while ((index > 0) && (heap[parent].Priority < he.Priority)) { heap[index] = heap[parent]; index = parent; parent = (index - 1) / 2; } heap[index] = he; } } private void trickleDown(int index, HeapEntry he) { int child = (index * 2) + 1; while (child < count) { if (this.ascend) { if (((child + 1) < count) && (heap[child].Priority > heap[child + 1].Priority)) { child++; } } else { if (((child + 1) < count) && (heap[child].Priority < heap[child + 1].Priority)) { child++; } } heap[index] = heap[child]; index = child; child = (index * 2) + 1; } bubbleUp(index, he); } } }
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