LeetCode 2233. 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

如何让乘积最大，这里面数组里面的数越平均，得到的值就越大，所以我们每次加1的时候，加在最小的那个数上面就可以，所以我们就用优先队列来解决这个问题就可以了

下面是代码：

Runtime: 456 ms, faster than 12.87% of Java online submissions for Maximum Product After K Increments.

Memory Usage: 53.3 MB, less than 58.48% of Java online submissions for Maximum Product After K Increments.