What is the time complexity of Find Minimum Operations to Make All Elements Divisible by Three?

The optimal solution for Find Minimum Operations to Make All Elements Divisible by Three runs in O(n) time and uses O(1) space. The time complexity remains O(n) as we still iterate through the array once, and the space complexity is O(1) since we use a constant amount of extra space.

What is the brute force solution for Find Minimum Operations to Make All Elements Divisible by Three?

The brute force approach for Find Minimum Operations to Make All Elements Divisible by Three is Brute Force, with O(n) time complexity and O(1) space complexity. The brute force approach involves checking each number in the array and determining how many operations are needed to make it divisible by 3. This method is straightforward but inefficient, as it may involve unnecessary calculations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Minimum Operations to Make All Elements Divisible by Three?

Find Minimum Operations to Make All Elements Divisible by Three uses the following concepts: Array, Math. The recommended patterns to study are: Math, Array.

What are the interview tips for Find Minimum Operations to Make All Elements Divisible by Three?

Always think about edge cases, such as numbers that are already divisible by 3. Try to simplify the problem by looking for patterns in the operations needed. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Find Minimum Operations to Make All Elements Divisible by Three?

Not considering the case where the number is already divisible by 3. Overcomplicating the logic by trying to handle each number differently instead of using a uniform approach.

#3190

Find Minimum Operations to Make All Elements Divisible by Three

Easy

ArrayMathMathArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution leverages the fact that each number can be adjusted to the nearest multiple of 3 in at most one operation. By calculating the necessary adjustments directly, we can achieve the result more efficiently.

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Algorithm

3 steps

1Step 1: Initialize a counter for operations to 0.
2Step 2: For each element in the array, calculate the remainder when divided by 3.
3Step 3: If the remainder is not zero, add the minimum of the remainder and (3 - remainder) to the operations counter.

solution.py2 lines

1def min_operations(nums):
2    return sum(min(num % 3, 3 - (num % 3)) for num in nums)

ℹ

Complexity note: The time complexity remains O(n) as we still iterate through the array once, and the space complexity is O(1) since we use a constant amount of extra space.

1Each number can be adjusted to a multiple of 3 in at most one operation.
2Understanding the modulo operation helps in determining the necessary adjustments.

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