Introduction

In this blog post, we will explore how to solve the LeetCode problem "1262" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an integer array nums, return the maximum possible sum of elements of the array such that it is divisible by three. Example 1: Input: nums = [3,6,5,1,8] Output: 18 Explanation: Pick numbers 3, 6, 1 and 8 their sum is 18 (maximum sum divisible by 3). Example 2: Input: nums = [4] Output: 0 Explanation: Since 4 is not divisible by 3, do not pick any number. Example 3: Input: nums = [1,2,3,4,4] Output: 12 Explanation: Pick numbers 1, 3, 4 and 4 their sum is 12 (maximum sum divisible by 3). Constraints: 1 <= nums.length <= 4 * 104 1 <= nums[i] <= 104

Explanation

Here's a solution to the problem, designed for efficiency and optimality:

• Dynamic Programming: Use dynamic programming to track the maximum sum achievable for each possible remainder (0, 1, and 2) when dividing by 3.
• Iterative Update: Iterate through the input array, updating the DP table at each step to reflect the inclusion of the current number in the sum.

• Modulo Arithmetic: Efficiently calculate remainders using the modulo operator (%) to determine the appropriate DP table updates.

• Runtime Complexity: O(n), where n is the length of the input array. Storage Complexity: O(1) (constant space).

Code

    def maxSumDivisibleByThree(nums):
    """
    Finds the maximum sum of elements in the array that is divisible by three.

    Args:
        nums: A list of integers.

    Returns:
        The maximum sum divisible by three.
    """

    dp = [0, float('-inf'), float('-inf')]  # dp[i] is max sum with remainder i

    for num in nums:
        temp_dp = dp[:]  # Create a copy to avoid overwriting previous states
        remainder = num % 3

        dp[0] = max(temp_dp[0], temp_dp[(3 - remainder) % 3] + num)
        dp[1] = max(temp_dp[1], temp_dp[(4 - remainder) % 3] + num)
        dp[2] = max(temp_dp[2], temp_dp[(5 - remainder) % 3] + num)

    return dp[0]