<p>You are given an array of non-negative integers <code>nums</code> and an integer <code>k</code>. In one operation, you may choose <strong>any</strong> element from <code>nums</code> and <strong>increment</strong> it by <code>1</code>.</p>

<p>Return<em> the <strong>maximum</strong> <strong>product</strong> of </em><code>nums</code><em> after <strong>at most</strong> </em><code>k</code><em> operations. </em>Since the answer may be very large, return it <b>modulo</b> <code>10<sup>9</sup> + 7</code>. Note that you should maximize the product before taking the modulo.&nbsp;</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,4], k = 5
<strong>Output:</strong> 20
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [6,3,3,2], k = 2
<strong>Output:</strong> 216
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length, k &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution:
    def maximumProduct(self, nums: List[int], k: int) -> int:
        mod = pow(10, 9) + 7
        n = len(nums)
        if n == 1:
            return nums[0] + k % mod
        product = 1
        pq = []
        # 对任意nums[i]执行k次加1操作, 求整个nums的最大可能乘积
        # 等同短板问题, 将k个短板都加1, 实际上是求k-minheap. O(nlogk)
        # if n >= k:
        for num in nums:
            heapq.heappush(pq, num)
        while k > 0:
            e1 = heapq.heappop(pq)
            e2 = heapq.heappop(pq)
            diff = min(e2 - e1 + 1, k)
            e1 += diff
            k -= diff
            heapq.heappush(pq, e1)
            heapq.heappush(pq, e2)
        
        for e in pq:
            product = (product * e) % mod
        return product
1265