How do you solve Maximum Sum of Three Numbers Divisible by Three on LeetCode?

Maximum Sum of Three Numbers Divisible by Three (LeetCode #3780) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Divisibility by 3 depends on the sum of remainders.

What is the time complexity of Maximum Sum of Three Numbers Divisible by Three?

The optimal solution for Maximum Sum of Three Numbers Divisible by Three runs in O(n log n) time and uses O(n) space. Sorting the remainder lists takes O(n log n), while the rest is linear.

What is the brute force solution for Maximum Sum of Three Numbers Divisible by Three?

The brute force approach for Maximum Sum of Three Numbers Divisible by Three is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all combinations of three numbers to find the maximum sum divisible by three. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Sum of Three Numbers Divisible by Three?

Maximum Sum of Three Numbers Divisible by Three uses the following concepts: Array, Greedy, Sorting, Heap (Priority Queue). The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Sum of Three Numbers Divisible by Three?

Explain your thought process clearly. Consider edge cases before finalizing your solution. Practice similar problems to reinforce your understanding.

What are common mistakes when solving Maximum Sum of Three Numbers Divisible by Three?

Not considering all combinations of remainders. Overlooking the case where no valid triplet exists.

#3780

Maximum Sum of Three Numbers Divisible by Three

Medium

ArrayGreedySortingHeap (Priority Queue)Hash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(n)

Group numbers by their remainder when divided by three, then select combinations that yield the maximum sum divisible by three.

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Algorithm

3 steps

1Step 1: Create three lists for numbers with remainders 0, 1, and 2 when divided by 3.
2Step 2: Sort each list in descending order to prioritize larger numbers.
3Step 3: Calculate potential sums using combinations of numbers from these lists and return the maximum valid sum.

solution.py11 lines

1def maxSumDivThree(nums):
2    rem = [[], [], []]
3    for num in nums:
4        rem[num % 3].append(num)
5    for r in rem:
6        r.sort(reverse=True)
7    max_sum = 0
8    for i in range(3):
9        if len(rem[i]) >= 3:
10            max_sum = max(max_sum, sum(rem[i][:3]))
11    return max_sum

ℹ

Complexity note: Sorting the remainder lists takes O(n log n), while the rest is linear.

1Divisibility by 3 depends on the sum of remainders.
2Sorting helps in maximizing the sum efficiently.

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