How do you solve Maximum Product After K Increments on LeetCode?

Maximum Product After K Increments (LeetCode #2233) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + k log n) time complexity and O(n) space complexity. Key insight: Always prioritize incrementing the smallest number to maximize the product.

What is the brute force solution for Maximum Product After K Increments?

The brute force approach for Maximum Product After K Increments is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying every possible way to increment the elements of the array and calculating the product after each increment. This method is straightforward but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n + k log n).

What algorithm or data structure is used to solve Maximum Product After K Increments?

Maximum Product After K Increments uses the following concepts: Array, Greedy, Heap (Priority Queue). The recommended patterns to study are: Heap (Priority Queue), Greedy Algorithms.

What are the interview tips for Maximum Product After K Increments?

Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Be prepared to discuss the trade-offs between different approaches. Practice coding heap operations, as they are commonly used in optimization problems.

What are common mistakes when solving Maximum Product After K Increments?

Not considering the impact of incrementing larger numbers first, which can lead to suboptimal products. Failing to use efficient data structures like heaps when dealing with dynamic minimum retrieval.

#2233

Maximum Product After K Increments

Medium

ArrayGreedyHeap (Priority Queue)Heap (Priority Queue)Greedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + k log n)
Space	O(1)	O(n)

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Intuition

Time O(n + k log n)Space O(n)

The optimal approach uses a priority queue (min-heap) to always increment the smallest number in the array. This ensures that each increment has the maximum possible effect on the product.

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Algorithm

3 steps

1Step 1: Create a min-heap from the nums array to efficiently access the smallest element.
2Step 2: For k times, extract the smallest element, increment it by 1, and push it back into the heap.
3Step 3: After all increments, calculate the product of all elements in the heap.

solution.py13 lines

1# Full working Python code
2import heapq
3
4def maxProduct(nums, k):
5    heapq.heapify(nums)
6    for _ in range(k):
7        smallest = heapq.heappop(nums)
8        smallest += 1
9        heapq.heappush(nums, smallest)
10    product = 1
11    for num in nums:
12        product = (product * num) % (10**9 + 7)
13    return product

ℹ

Complexity note: The time complexity is dominated by the heap operations, where each increment takes log n time, and we perform this k times. The space complexity is due to storing the heap.

1Always prioritize incrementing the smallest number to maximize the product.
2Using a min-heap allows for efficient retrieval and updating of the smallest element.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.