O(nlogn) time and O(n) space using min heap

Author: 149ps

2233. Maximum Product After K Increments/2233. Maximum Product After K Increments.py

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+"""
+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.
+
+
+
+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
+"""
+class Solution:
+    def maximumProduct(self, nums: List[int], k: int) -> int:
+        if len(nums) == 1: return nums[0] + k
+        min_heap = nums
+        heapq.heapify(min_heap)
+        while k > 0:
+            current = heapq.heappop(min_heap)
+            cur_diff = min_heap[0] - current
+            if cur_diff == 0:
+                current += 1
+                k -= 1
+            elif cur_diff > k:
+                current += k
+                k = 0
+            else:
+                current += cur_diff
+                k -= cur_diff
+            heapq.heappush(min_heap,current)
+        result = 1
+        while len(min_heap) > 0:
+            cur = heapq.heappop(min_heap)
+            result = (result * cur) % (10**9+7)
+        return result