Solution: Maximum Product After K Increments
Let’s solve the Maximum Product After K Increments using the Top K Elements pattern.
We'll cover the following
Statement
You are given an array, nums
, consisting of non-negative integers, and an integer k
representing the maximum number of allowed operations.
In each operation, you may select any element in nums
and increment it by k
such operations.
Your task is to maximize the product of all elements in the array after performing up to k
operations. As the resulting product can be very large, return the product modulo
Note: Ensure that the product is maximized before applying the modulo operation.
Constraints:
nums.length
,k
nums[i]
Solution
The core intuition behind the solution is to maximize the product of the nums
elements by distributing the k
increment operations in a way that promotes balance among the values. In multiplication, products are higher when numbers are close in value. For example,
Multiplication favors numbers that are close in value, so the goal is to balance out
nums
by incrementing the smallest elements first.Incrementing smaller numbers contributes more to the overall product than incrementing larger ones, making them the most effective targets for
k
operations.
This strategy uses the top k elements pattern with a min heap, allowing quick access to the current smallest number. All elements from nums
are inserted into the heap. For each k
operations, we extract the minimum element, increment it by
In some cases, the smallest element may be much smaller than the rest of the array—for example, in k
operations may be applied solely to that minimum element, progressively closing the gap and improving the product.
After performing all k
increments, the product of all heap elements is computed and returned modulo
Now, let’s look at the solution steps below:
We initialize a variable
mod = 10
9
+ 7
to handle modulo operations for the overflow issues.We initialize a min heap with all the elements of
nums
.To evenly distribute increments, we perform
k
operations by:Popping the
smallest
element from the heap.Incrementing the
smallest
element by. Pushing the
smallest
element back into the heap.
We initialize a variable,
result = 1
, to store the maximized product of all elements innums
.After
k
increments, we compute the product modulo of all elements in the heap by:Repeatedly popping elements from the heap.
Multiply them with the
result
, taking the modulo withmod
at each step to prevent overflow.
We return the final value of the
result
, which represents the maximum product modulo.
Let’s look at the following illustration to get a better understanding of the solution:
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