How do you solve Greatest Sum Divisible by Three on LeetCode?

Greatest Sum Divisible by Three (LeetCode #1262) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using dynamic programming allows us to efficiently track potential sums without generating all subsets.

What is the brute force solution for Greatest Sum Divisible by Three?

The brute force approach for Greatest Sum Divisible by Three is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible subsets of the array to find the maximum sum that is divisible by three. This method is straightforward but inefficient for larger arrays due to the exponential number of subsets. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Greatest Sum Divisible by Three?

Greatest Sum Divisible by Three uses the following concepts: Array, Dynamic Programming, Greedy, Sorting. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Greatest Sum Divisible by Three?

Always clarify the problem and constraints before jumping into coding. Think about edge cases and how they might affect your solution. Explain your thought process clearly while coding, as it helps the interviewer understand your approach.

What are common mistakes when solving Greatest Sum Divisible by Three?

Failing to consider edge cases where no elements can contribute to a sum divisible by 3. Not properly updating the state in dynamic programming, leading to incorrect results.

#1262

Greatest Sum Divisible by Three

Medium

ArrayDynamic ProgrammingGreedySortingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

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)

The optimal solution uses dynamic programming to keep track of the maximum sums that can be achieved with different remainders when divided by 3. This allows us to efficiently compute the maximum sum without checking all subsets.

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Algorithm

3 steps

1Step 1: Initialize an array dp of size 3 with all zeros to store maximum sums for remainders 0, 1, and 2.
2Step 2: Iterate through each number in the input array.
3Step 3: For each number, update the dp array based on the current number's contribution to each remainder.

solution.py8 lines

1def maxSumDivThree(nums):
2    dp = [0, 0, 0]
3    for num in nums:
4        temp = dp[:]
5        for i in range(3):
6            temp[(i + num) % 3] = max(temp[(i + num) % 3], dp[i] + num)
7        dp = temp
8    return dp[0]

ℹ

Complexity note: The time complexity is O(n) because we iterate through the array once, and the space complexity is O(1) since we only use a fixed-size array to store the maximum sums.

1Using dynamic programming allows us to efficiently track potential sums without generating all subsets.
2Understanding remainders when dividing by 3 helps in optimizing the solution.

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