#2233

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Maximum Product After K Increments

Maximize the product of an array after performing up to k increments using a greedy approach with heap optimization.

Problem Statement

Given an array of non-negative integers nums and an integer k, you may increment any element by 1 in each of k operations. Return the maximum possible product of nums after at most k operations, modulo 10^9+7.

For example, nums = [0,4] and k = 5 allows incrementing 0 five times to get [5,4], producing the maximum product 20. You must find the strategy that maximizes the product before applying the modulo.

Examples

Example 1

Input: nums = [0,4], k = 5

Output: 20

Increment the first number 5 times. Now nums = [5, 4], with a product of 5 * 4 = 20. It can be shown that 20 is maximum product possible, so we return 20. Note that there may be other ways to increment nums to have the maximum product.

Example 2

Input: nums = [6,3,3,2], k = 2

Output: 216

Increment the second number 1 time and increment the fourth number 1 time. Now nums = [6, 4, 3, 3], with a product of 6 * 4 * 3 * 3 = 216. It can be shown that 216 is maximum product possible, so we return 216. Note that there may be other ways to increment nums to have the maximum product.

Constraints

1 <= nums.length, k <= 105
0 <= nums[i] <= 106

Solution Approach

Use a Min-Heap for Greedy Increments

Push all numbers into a min-heap to quickly access the smallest element. Increment the minimal element in each operation, then re-insert it into the heap to maintain the heap property. This ensures that each increment contributes most to the final product.

Maintain Modulo While Computing Product

After completing all k increments, pop elements from the heap and multiply them together while applying modulo 10^9+7 at each multiplication. This prevents integer overflow and aligns with problem constraints.

Early Optimization for Large k

If k is larger than the number of elements, batch increments by dividing k among the smallest elements efficiently. This reduces the number of heap operations while still maintaining the greedy choice principle.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n log n + k log n) due to heap operations for each increment. Space complexity is O(n) for the heap storage.

What Interviewers Usually Probe

Expect a solution that maximizes product using minimal increments first.
Using a heap to track the smallest number indicates awareness of greedy invariant.
Be ready to discuss modulo arithmetic and why it's applied during multiplication.

Common Pitfalls or Variants

Common pitfalls

Incrementing larger numbers first can reduce the final product and violates greedy choice.
Not using modulo while multiplying can cause integer overflow.
Failing to maintain the heap invariant after each increment results in suboptimal operations.

Follow-up variants

Find maximum sum instead of product after k increments using a similar greedy approach.
Allow decrement operations as well and maximize product using mixed greedy decisions.
Compute maximum product if increments must be applied to consecutive elements only.

FAQ

What is the main pattern used in Maximum Product After K Increments?

The main pattern is greedy choice plus invariant validation, incrementing the smallest element repeatedly to maximize the product.

Why is a heap useful for this problem?

A min-heap efficiently tracks the smallest number, ensuring each increment contributes maximally to the product, maintaining the greedy invariant.

How do we handle large products exceeding integer limits?

Apply modulo 10^9+7 during multiplication to prevent overflow while keeping the correct result.

Can we increment multiple elements at once?

Yes, but to maintain the greedy principle, increments should prioritize the current minimal elements first.

What is a common mistake with k increments in this problem?

A frequent error is distributing increments without regard for element size, which can lead to a lower final product.

terminal

Solution

Solution 1: Greedy + Priority Queue (Min-Heap)

According to the problem description, to maximize the product, we need to increase the smaller numbers as much as possible. Therefore, we can use a min-heap to maintain the array $\textit{nums}$. Each time, we take the smallest number from the min-heap, increase it by $1$, and then put it back into the min-heap. After repeating this process $k$ times, we multiply all the numbers currently in the min-heap to get the answer.

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class Solution:
    def maximumProduct(self, nums: List[int], k: int) -> int:
        heapify(nums)
        for _ in range(k):
            heapreplace(nums, nums[0] + 1)
        mod = 10**9 + 7
        return reduce(lambda x, y: x * y % mod, nums)

Continue Practicing

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