Problem Description

You have an array of integers nums (0-indexed) and an integer k. You need to perform exactly k operations to maximize your score.

Each operation consists of:

1. Select any element m from the array nums
2. Remove that element from the array
3. Add a new element with value m + 1 back to the array
4. Add m to your score

Your goal is to find the maximum possible score after performing exactly k operations.

For example, if you have nums = [1, 2, 3] and k = 2:

• First operation: You could select 3, remove it, add 4 to the array (making it [1, 2, 4]), and gain 3 points
• Second operation: You could select 4, remove it, add 5 to the array (making it [1, 2, 5]), and gain 4 points
• Total score: 3 + 4 = 7

The key insight is that to maximize the score, you should always select the largest element available in each operation. Since each operation replaces an element with a larger one (m becomes m + 1), repeatedly selecting the maximum element ensures you're always choosing the highest possible value. This leads to selecting values x, x + 1, x + 2, ..., x + k - 1 where x is the initial maximum element, giving a total score of k * x + (0 + 1 + 2 + ... + (k-1)) = k * x + k * (k - 1) / 2.

Intuition

To maximize our score, we need to think about which element to select in each operation. Since we're adding the value of the selected element to our score, we want to pick the largest possible element each time.

Let's think about what happens during the operations:

• When we select an element m, we remove it and add m + 1 back to the array
• This means we're essentially replacing m with m + 1
• The key observation is that this replacement creates an even larger element for potential future selections

Since we want to maximize our score, the greedy strategy is to always pick the maximum element available. But here's the clever part: if we start by picking the maximum element x from the original array, after the operation we'll have x + 1 in the array. If we pick x + 1 next time, we'll get x + 2, and so on.

This pattern reveals that if we consistently pick the maximum element, we'll be selecting:

• First operation: select x (the original max), score += x
• Second operation: select x + 1, score += x + 1
• Third operation: select x + 2, score += x + 2
• ...
• k-th operation: select x + k - 1, score += x + k - 1

The total score becomes: x + (x + 1) + (x + 2) + ... + (x + k - 1)

This can be simplified to: k * x + (0 + 1 + 2 + ... + (k - 1))

Using the arithmetic series formula, 0 + 1 + 2 + ... + (k - 1) = k * (k - 1) / 2

Therefore, the maximum score is: k * x + k * (k - 1) / 2

This mathematical insight allows us to calculate the answer directly without simulating the operations, making the solution extremely efficient.

Learn more about Greedy patterns.

Solution Approach

The implementation is straightforward once we understand the mathematical pattern. We use a greedy approach combined with mathematical optimization to solve this problem in constant time after finding the maximum element.

Step-by-step implementation:

1. Find the maximum element: First, we need to identify the largest element x in the array using max(nums). This takes O(n) time where n is the length of the array.

2. Apply the mathematical formula: Instead of simulating k operations, we directly calculate the result using the formula we derived:

   • The sum of selecting x, x+1, x+2, ..., x+k-1 is equivalent to k * x + (0 + 1 + 2 + ... + (k-1))
   • The arithmetic series sum 0 + 1 + 2 + ... + (k-1) equals k * (k - 1) / 2

3. Return the result: The final answer is k * x + k * (k - 1) // 2

Why this works:

• By always selecting the maximum element available, we ensure each operation contributes the maximum possible value to our score
• After selecting maximum element x and replacing it with x+1, the new maximum becomes x+1
• This pattern continues, creating an ascending sequence of selections: x, x+1, x+2, ..., x+k-1
• Since we know exactly what values we'll select, we can calculate the sum directly without simulation

Time Complexity: O(n) - We only need to find the maximum element once Space Complexity: O(1) - We only use a constant amount of extra space

The beauty of this solution lies in recognizing that the greedy choice (always pick the maximum) leads to a predictable pattern that can be calculated mathematically, avoiding the need for any actual array modifications or iterations through k operations.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example: nums = [5, 1, 3, 9, 2], k = 3

Step 1: Find the maximum element

• Scanning through the array: [5, 1, 3, 9, 2]
• Maximum element x = 9

Step 2: Understand what values we'll select Instead of actually performing the operations, we know that by always picking the maximum:

• Operation 1: Select 9, remove it, add 10 to array, score += 9
• Operation 2: Select 10 (new max), remove it, add 11 to array, score += 10
• Operation 3: Select 11 (new max), remove it, add 12 to array, score += 11

So we're selecting the sequence: 9, 10, 11

Step 3: Apply the mathematical formula Rather than simulating these operations, we calculate directly:

• We're summing: 9 + 10 + 11
• This equals: k * x + (0 + 1 + 2) where x = 9 and k = 3
• Which is: 3 * 9 + (0 + 1 + 2)
• Simplifying: 27 + 3
• Using the formula: k * x + k * (k - 1) / 2 = 3 * 9 + 3 * 2 / 2 = 27 + 3 = 30

Step 4: Verify with manual calculation Let's verify our formula gives the same result as adding the values directly:

• Direct sum: 9 + 10 + 11 = 30 ✓
• Formula result: 3 * 9 + 3 * (3 - 1) / 2 = 27 + 3 = 30 ✓

Final Answer: 30

The key insight is that we don't need to modify the array at all. Once we identify the maximum element (9), we can immediately calculate that selecting it k times (with it incrementing each time) gives us the maximum possible score of 30.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the length of the input array nums. The algorithm finds the maximum element in the array using max(nums), which requires a single pass through all elements. The arithmetic operations k * x + k * (k - 1) // 2 are performed in constant time O(1).

Space Complexity: O(1). The algorithm only uses a constant amount of extra space to store the maximum value x and perform the calculation. No additional data structures that scale with input size are created.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow with Large Values

When dealing with large values of k and max_element, the multiplication k * max_element or the formula k * (k - 1) // 2 might cause integer overflow in languages with fixed integer sizes. While Python handles arbitrary precision integers automatically, this could be problematic in other languages.

Solution: In languages like Java or C++, use appropriate data types (e.g., long long in C++ or long in Java) to handle large results. Always validate the constraints to ensure the result fits within the expected range.

2. Misunderstanding the Operation Mechanics

A common mistake is thinking that we need to track or modify the actual array during operations. Some might implement a simulation approach using a heap or priority queue to repeatedly extract and insert elements, which would be unnecessarily complex and inefficient.

Incorrect approach example:

# Inefficient simulation approach
import heapq
max_heap = [-x for x in nums]  # Convert to max heap
heapq.heapify(max_heap)
score = 0
for _ in range(k):
    max_val = -heapq.heappop(max_heap)
    score += max_val
    heapq.heappush(max_heap, -(max_val + 1))
return score

Solution: Recognize that the greedy pattern leads to a mathematical formula. Since we always pick the maximum and it increases by 1 each time, we can calculate the result directly without any simulation.

3. Off-by-One Errors in the Arithmetic Series

When calculating the sum of the arithmetic series, it's easy to make mistakes with the formula. Some might incorrectly use k * (k + 1) // 2 instead of k * (k - 1) // 2, forgetting that we're summing from 0 to k-1, not 1 to k.

Solution: Remember that we're adding the increments: 0, 1, 2, ..., (k-1). The sum of first n natural numbers starting from 0 is n * (n - 1) / 2, not n * (n + 1) / 2.

4. Empty Array or k = 0 Edge Cases

While the problem likely guarantees non-empty arrays and positive k, not handling these edge cases could cause runtime errors.

Solution: Add validation if needed:

if not nums or k == 0:
    return 0

5. Floating Point Division Instead of Integer Division

Using regular division / instead of integer division // might introduce floating point numbers where integers are expected, potentially causing type errors or precision issues.

Solution: Always use integer division // when dealing with integer arithmetic to ensure the result remains an integer.