Leetcode 2530: Maximal Score After Applying K Operations
You are given a list of integers
nums and an integer k. In each operation, you choose an index i, increase your score by nums[i], and replace nums[i] with ceil(nums[i] / 3). Apply exactly k operations and return the maximum score you can achieve.Problem
Approach
Steps
Complexity
Input: The input consists of an array `nums` of integers and an integer `k`.
Example: [10, 10, 10, 10, 10], 5
Constraints:
• 1 <= nums.length, k <= 10^5
• 1 <= nums[i] <= 10^9
Output: The output is a single integer representing the maximum score after applying exactly `k` operations.
Example: 50
Constraints:
• The output will be an integer representing the maximum score achievable.
Goal: The goal is to maximize the score by carefully selecting the index `i` at each operation.
Steps:
• Initialize a max-heap (priority queue) with all elements of `nums`.
• For each of the `k` operations, extract the maximum element, increase the score by this value, and replace it with `ceil(nums[i] / 3)`.
• Repeat the operation for exactly `k` times and return the final score.
Goal: The length of `nums` and `k` are at most 100,000, and each element of `nums` can be as large as 10^9.
Steps:
• 1 <= nums.length, k <= 10^5
• 1 <= nums[i] <= 10^9
Assumptions:
• You must apply exactly `k` operations.
• The `nums` array is modified after each operation.
• Input: [15, 10, 20, 5, 30], 5
• Explanation: In this example, after applying the operation to each element, the final score is 15 + 20 + 30 + 5 + 10 = 80.
• Input: [10, 20, 15, 10], 3
• Explanation: The optimal sequence of operations results in a final score of 45.
• Input: [1, 1, 1, 1, 1], 3
• Explanation: In this case, after 3 operations, the score is simply 3.
Approach: To maximize the score, always choose the largest element for each operation, ensuring the most valuable elements are reduced in value last.
Observations:
• Using a max-heap (priority queue) helps efficiently find the largest element for each operation.
• The solution needs to be efficient, as `nums.length` and `k` can be as large as 100,000.
Steps:
• Initialize a max-heap with the elements of `nums`.
• Perform exactly `k` operations: extract the max element, add it to the score, and push `ceil(nums[i] / 3)` back into the heap.
• Return the final score after `k` operations.
Empty Inputs:
• The array `nums` is always non-empty, as the length is at least 1.
Large Inputs:
• Ensure that the solution works efficiently for large inputs (up to 100,000 elements and operations).
Special Values:
• Handle cases where elements are small, such as 1, which will always be replaced by 1 after an operation.
Constraints:
• The problem requires an efficient approach, such as using a max-heap to handle up to 100,000 operations efficiently.
long long maxKelements(vector<int>& nums, int k) {
priority_queue<int> pq;
for(int i = 0; i < nums.size(); i++) {
pq.push(nums[i]);
}
long long ans = 0;
while(k--) {
int tmp = pq.top();
cout << tmp << " ";
ans += tmp;
pq.pop();
pq.push(ceil(tmp/3.0));
}
return ans;
}
1 : Function Definition
long long maxKelements(vector<int>& nums, int k) {
This is the function definition for 'maxKelements', which accepts a vector of integers 'nums' and an integer 'k', returning a long long value.
2 : Variable Initialization
priority_queue<int> pq;
A priority queue 'pq' is declared to store integers. This will be used to efficiently access the largest element at any time.
3 : Loop
for(int i = 0; i < nums.size(); i++) {
A for loop is initiated to iterate over the vector 'nums' and push each element into the priority queue.
4 : Queue Operation
pq.push(nums[i]);
Each element from the 'nums' vector is pushed into the priority queue 'pq', which will automatically arrange the elements in descending order.
5 : Variable Initialization
long long ans = 0;
A variable 'ans' is initialized to 0 to store the cumulative sum of the largest elements selected from the queue.
6 : Loop
while(k--) {
A while loop runs 'k' times to extract the largest element, update it, and accumulate the result.
7 : Queue Operation
int tmp = pq.top();
The top element of the priority queue (the largest element) is stored in the variable 'tmp'.
8 : Accumulate Sum
ans += tmp;
The value of 'tmp' is added to the cumulative sum 'ans'.
9 : Queue Operation
pq.pop();
The top element (which has already been used) is removed from the priority queue.
10 : Queue Operation
pq.push(ceil(tmp/3.0));
The value 'tmp' is divided by 3 and rounded up, then pushed back into the priority queue to be used in subsequent iterations.
11 : Return Statement
return ans;
The final accumulated sum 'ans' is returned as the result of the function.
Best Case: O(k log n)
Average Case: O(k log n)
Worst Case: O(k log n)
Description: Each heap operation (extract and insert) takes O(log n) time, and we perform `k` operations.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the storage of the heap, where `n` is the length of the `nums` array.
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