What is the time complexity of Minimum Operations to Make Array Sum Divisible by K?

The optimal solution for Minimum Operations to Make Array Sum Divisible by K runs in O(n) time and uses O(1) space. The sum calculation is O(n), and we only need a constant amount of space.

What is the brute force solution for Minimum Operations to Make Array Sum Divisible by K?

The brute force approach for Minimum Operations to Make Array Sum Divisible by K is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try reducing each element in the array until the sum becomes divisible by k. This involves checking every possible combination, which can be inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Operations to Make Array Sum Divisible by K?

Minimum Operations to Make Array Sum Divisible by K uses the following concepts: Array, Math. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Operations to Make Array Sum Divisible by K?

Understand the problem requirements clearly. Practice identifying patterns in array problems. Focus on optimizing brute force solutions to improve efficiency.

What are common mistakes when solving Minimum Operations to Make Array Sum Divisible by K?

Not checking if the sum is already divisible by k. Assuming you need to reduce multiple elements instead of just calculating the remainder.

#3512

Minimum Operations to Make Array Sum Divisible by K

Easy

ArrayMathHash MapArray

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

Instead of reducing elements one by one, we can calculate how much we need to reduce the total sum to make it divisible by k, which is more efficient.

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Algorithm

3 steps

1Step 1: Calculate the sum of the array.
2Step 2: Find the remainder when the sum is divided by k.
3Step 3: The minimum operations needed is equal to the remainder, since we can reduce any element directly to achieve divisibility.

solution.py4 lines

1def min_operations(nums, k):
2    total = sum(nums)
3    remainder = total % k
4    return remainder

ℹ

Complexity note: The sum calculation is O(n), and we only need a constant amount of space.

1The sum's remainder when divided by k directly gives the number of operations needed.
2Reducing elements one by one is inefficient compared to calculating the total sum's remainder.

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