Leetcode #2233: Maximum Product After K Increments
In this guide, we solve Leetcode #2233 Maximum Product After K Increments in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an array of non-negative integers nums and an integer k. In one operation, you may choose any element from nums and increment it by 1.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Array, Heap (Priority Queue)
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: nums = [0,4], k = 5
Output: 20
Explanation: 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.
Python Solution
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)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.