2530. Maximal Score After Applying K Operations - Detailed Explanation

Free Coding Questions Catalog

Boost your coding skills with our essential coding questions catalog. Take a step towards a better tech career now!

Problem Statement

Description:
You are given an array of positive integers nums and an integer k. You must perform exactly k operations. In each operation, you:

Choose the maximum element from nums.
Add its value to your total score.
Replace that element with its floor division by 2 (i.e. update it to ⌊x/2⌋).

Return the maximum score you can obtain after performing k operations.

Example Inputs & Outputs:

Example 1:
- Input:
  - nums = [10, 20, 7]
  - k = 4
- Process:
  - Operation 1:
    - Maximum is 20.
    - Add 20 to score (score = 20).
    - Update 20 → 20 // 2 = 10.
    - New array: [10, 10, 7].
  - Operation 2:
    - Maximum is 10 (one of them).
    - Add 10 to score (score = 30).
    - Update 10 → 10 // 2 = 5.
    - New array: [5, 10, 7].
  - Operation 3:
    - Maximum is 10.
    - Add 10 to score (score = 40).
    - Update 10 → 10 // 2 = 5.
    - New array: [5, 5, 7].
  - Operation 4:
    - Maximum is 7.
    - Add 7 to score (score = 47).
    - Update 7 → 7 // 2 = 3.
    - New array becomes [5, 5, 3].
- Output: 47
Example 2:
- Input:
  - nums = [5, 3, 8]
  - k = 3
- Process:
  - Operation 1: Maximum is 8; score = 8; update 8 → 4; array: [5, 3, 4].
  - Operation 2: Maximum is 5; score = 8 + 5 = 13; update 5 → 2; array: [2, 3, 4].
  - Operation 3: Maximum is 4; score = 13 + 4 = 17; update 4 → 2; array: [2, 3, 2].
- Output: 17

Constraints:

$(1 \leq \text{nums.length} \leq 10^5)$ (or similar, depending on the variant)
$(1 \leq k \leq 10^5)$
$(1 \leq \text{nums}[i] \leq 10^9)$

Hints to Approach the Problem

Hint 1:
In every operation, the best strategy is to add the largest available number to your score.
Hint 2:
After you choose the maximum element, its contribution in future operations is reduced (since you replace it with its half). Thus, you need a data structure that always gives you the maximum quickly and lets you update values efficiently.
Hint 3:
A max‑heap (or priority queue with a reverse order) is ideal because it lets you extract the maximum element in $(O(\log n))$ time and update the heap in $(O(\log n))$ time.

Approaches

Approach: Greedy Using a Max‑Heap

Idea:
Always pick the current largest element (using a max‑heap). Add its value to your score, then push back its half (using floor division) into the heap. Repeat this process exactly k times.
Steps:
1. Initialize:
  Build a max‑heap from the array nums.
2. Iterate for k Operations:
  - Extract the maximum element from the heap.
  - Add this element to the total score.
  - Compute the new value as the floor division of the maximum element by 2.
  - Push this new value back into the heap.
3. Return the Total Score.
Why It’s Optimal:
By always choosing the largest available number, you ensure that each operation contributes as much as possible to the final score. The heap operations run in $(O(\log n))$ time, leading to an overall time complexity of $(O(k \log n))$ .

Python Code

Python3

. . . .

Java Code

Java

. . . .

Complexity Analysis

Time Complexity:
- Building the heap: (O(n)).
- Each of the (k) operations involves a heap extraction and insertion, each $(O(\log n))$ .
- Overall: $(O(n + k \log n))$ .
  
  (For large (k), the dominant term is $(O(k \log n))$ .)
Space Complexity:
- (O(n)) for the heap.

Step-by-Step Walkthrough and Visual Example

Let’s walk through Example 1 with nums = [10, 20, 7] and k = 4:

Initial Heap Construction:
- Convert array to max‑heap (using negatives in Python or a reverse comparator in Java):
  Heap content (as max‑heap): [20, 10, 7].
Operation 1:
- Extract Maximum: 20
- Score: (0 + 20 = 20)
- Update: 20 // 2 = 10
- Heap becomes: [10, 10, 7]
Operation 2:
- Extract Maximum: 10 (one of the 10s)
- Score: (20 + 10 = 30)
- Update: 10 // 2 = 5
- Heap becomes: [10, 7, 5]
Operation 3:
- Extract Maximum: 10
- Score: (30 + 10 = 40)
- Update: 10 // 2 = 5
- Heap becomes: [7, 5, 5]
Operation 4:
- Extract Maximum: 7
- Score: (40 + 7 = 47)
- Update: 7 // 2 = 3
- Heap becomes: [5, 5, 3]
Final Score: 47

Visual Summary:

Initial nums: [10, 20, 7]  → Heap: [20, 10, 7]
Op1: pop 20, score = 20, push 10  → Heap: [10, 10, 7]
Op2: pop 10, score = 30, push 5   → Heap: [10, 7, 5]
Op3: pop 10, score = 40, push 5   → Heap: [7, 5, 5]
Op4: pop 7,  score = 47, push 3   → Heap: [5, 5, 3]

Common Pitfalls

Not Using a Max‑Heap:
Using a min‑heap directly (or not adjusting for maximum extraction) will yield incorrect results.
Incorrect Floor Division:
Make sure to use integer division (or floor division) so that the value is updated correctly.
Not Handling Large Inputs Efficiently:
Even though the operations are simple, using inefficient data structures may lead to timeouts when (k) or (n) is large.

Edge Cases

Single Element Array:
When there is only one number, the same number will be chosen repeatedly.
k = 0:
If no operations are to be performed (if allowed by the constraints), then the score should remain 0.
Large Values:
Ensure that the data type used for the score can handle large sums (e.g., using a long type in Java).

Alternative Variations:
- Variants where the update operation is different (for example, subtracting a fixed value instead of dividing by 2).
- Maximizing or minimizing a different aggregate measure after a series of operations.
Related Problems for Further Practice:
- "Maximum Score From Performing Multiplication Operations" – A dynamic programming problem with similar greedy insights.
- "Merge Stones" – Where combining elements yields cumulative scores.
- "Minimize Deviation in Array" – Involves greedy operations on array elements.