1+ """
2+ 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.
3+
4+ Return the maximum product of nums after at most k operations. Since the answer may be very large, return it modulo 109 + 7.
5+
6+
7+
8+ Example 1:
9+
10+ Input: nums = [0,4], k = 5
11+ Output: 20
12+ Explanation: Increment the first number 5 times.
13+ Now nums = [5, 4], with a product of 5 * 4 = 20.
14+ It can be shown that 20 is maximum product possible, so we return 20.
15+ Note that there may be other ways to increment nums to have the maximum product.
16+ Example 2:
17+
18+ Input: nums = [6,3,3,2], k = 2
19+ Output: 216
20+ Explanation: Increment the second number 1 time and increment the fourth number 1 time.
21+ Now nums = [6, 4, 3, 3], with a product of 6 * 4 * 3 * 3 = 216.
22+ It can be shown that 216 is maximum product possible, so we return 216.
23+ Note that there may be other ways to increment nums to have the maximum product.
24+
25+
26+ Constraints:
27+
28+ 1 <= nums.length, k <= 105
29+ 0 <= nums[i] <= 106
30+ """
31+ class Solution :
32+ def maximumProduct (self , nums : List [int ], k : int ) -> int :
33+ if len (nums ) == 1 : return nums [0 ] + k
34+ min_heap = nums
35+ heapq .heapify (min_heap )
36+ while k > 0 :
37+ current = heapq .heappop (min_heap )
38+ cur_diff = min_heap [0 ] - current
39+ if cur_diff == 0 :
40+ current += 1
41+ k -= 1
42+ elif cur_diff > k :
43+ current += k
44+ k = 0
45+ else :
46+ current += cur_diff
47+ k -= cur_diff
48+ heapq .heappush (min_heap ,current )
49+ result = 1
50+ while len (min_heap ) > 0 :
51+ cur = heapq .heappop (min_heap )
52+ result = (result * cur ) % (10 ** 9 + 7 )
53+ return result
0 commit comments