{x}
blog image

Maximum Product After K Increments

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.

Return the maximum product of nums after at most k operations. Since the answer may be very large, return it modulo 109 + 7. Note that you should maximize the product before taking the modulo. 

 

Example 1:

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.

Example 2:

Input: nums = [6,3,3,2], k = 2
Output: 216
Explanation: 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 Explanation: Maximum Product After K Increments

This problem asks to find the maximum product of an array of non-negative integers after performing at most k increment operations on its elements. The key insight is a greedy approach: it's always better to increment the smallest numbers. This is because the multiplicative effect of increasing a small number is greater than increasing a larger one.

Algorithm:

  1. Use a Min-Heap: We utilize a min-heap data structure (priority queue) to efficiently track the smallest element in the array. Min-heaps allow us to quickly access and remove the minimum element.

  2. Iterative Increments: We iterate k times. In each iteration:

    • We extract the smallest element from the min-heap.
    • We increment the smallest element by 1.
    • We re-insert the incremented element back into the min-heap.

  3. Calculate Product: After k iterations, all increments are done. We then calculate the product of all elements in the min-heap, taking the modulo 10<sup>9</sup> + 7 to handle potential overflow.

Time Complexity:

  • Building the min-heap takes O(n) time, where n is the length of nums.
  • Each of the k iterations involves extracting the minimum (O(log n)), incrementing, and inserting back (O(log n)).
  • Calculating the final product takes O(n) time.

Therefore, the overall time complexity is O(n + k log n). If k is significantly smaller than n, this approaches O(n). If k is much larger than n, it's dominated by O(k log n).

Space Complexity:

The space complexity is dominated by the min-heap, which stores the n elements of the array. Therefore, the space complexity is O(n).

Code Examples (Python, Java, C++, Go, TypeScript, JavaScript):

The provided code snippets in multiple languages demonstrate this algorithm. They all follow the same structure:

  1. Initialize a Min-Heap: Create a min-heap data structure and populate it with the elements of nums.

  2. Increment Iteratively: Perform k iterations, extracting the minimum, incrementing, and re-inserting.

  3. Calculate Product (with modulo): Calculate the product of all elements in the heap, ensuring modulo operation to prevent integer overflow.

Example Usage (Python):

from heapq import heapify, heapreplace
import operator
from functools import reduce
 
nums = [0,4]
k = 5
solution = Solution()
result = solution.maximumProduct(nums,k) #result = 20
 
nums = [6,3,3,2]
k = 2
result = solution.maximumProduct(nums,k) #result = 216

The other languages follow a similar pattern, utilizing their respective priority queue implementations (e.g., PriorityQueue in Java, priority_queue in C++, container/heap in Go, MinPriorityQueue in TypeScript and JavaScript). The core logic remains the same.