What is the time complexity of Greatest Sum Divisible by Three?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Greatest Sum Divisible by Three?

This problem uses Array, Dynamic Programming, Greedy, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Greatest Sum Divisible by Three?

Insight 1: We only care about the remainders when dividing each number by 3. A number can only have remainders 0, 1, or 2 when divided by 3. Insight 2: Dynamic programming can be used to keep track of the maximum sum achievable for each possible remainder (0, 1, 2) when iterating through the array.

Which companies ask LeetCode 1262. Greatest Sum Divisible by Three in interviews?

This problem is frequently asked at DE Shaw.

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Greatest Sum Divisible by Three - LeetCode 1262 Solution

Greatest Sum Divisible by Three - Complete Solution Guide

Greatest Sum Divisible by Three is LeetCode problem 1262, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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). C

Detailed Explanation

The problem asks us to find the maximum possible sum of elements from a given integer array `nums` such that the sum is divisible by 3. We are allowed to pick any combination of numbers from the array. If no such combination exists (i.e., no subset sums to a multiple of 3), we should return 0.

Solution Approach

The solution uses dynamic programming to efficiently track the maximum possible sums for each remainder (0, 1, and 2) when dividing by 3. We initialize a `dp` array of size 3 with all elements set to 0. `dp[i]` will store the maximum possible sum seen so far with remainder `i` when divided by 3. We iterate through the input array `nums`. For each number, we update the `dp` array by considering including the current number in the sums and calculating the new remainders. At the end, `dp[0]` will contain the maximum possible sum that is divisible by 3.

Step-by-Step Algorithm

Step 1: Initialize a `dp` array of size 3 to [0, 0, 0]. `dp[i]` represents the maximum sum with remainder `i` when divided by 3.
Step 2: Iterate through each number `num` in the input array `nums`.
Step 3: Create a copy of the `dp` array (e.g., `dp_copy`) to avoid modifying it directly during the inner loop. This is crucial because updates to `dp` based on previous values in `dp` are performed.
Step 4: Iterate through each element `s` in `dp_copy`.
Step 5: Calculate the new sum: `new_sum = s + num`.
Step 6: Calculate the remainder when `new_sum` is divided by 3: `remainder = new_sum % 3`.
Step 7: Update `dp[remainder]` to the maximum of its current value and `new_sum`: `dp[remainder] = max(dp[remainder], new_sum)`.
Step 8: After iterating through all numbers in `nums`, return `dp[0]`, which represents the maximum sum divisible by 3.

Key Insights

Insight 1: We only care about the remainders when dividing each number by 3. A number can only have remainders 0, 1, or 2 when divided by 3.
Insight 2: Dynamic programming can be used to keep track of the maximum sum achievable for each possible remainder (0, 1, 2) when iterating through the array.
Insight 3: We don't need to store all intermediate sums; we only need to store the largest sum seen so far for each remainder (0, 1, 2).

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Greedy, Sorting.

Companies

Asked at: DE Shaw.

Greatest Sum Divisible by Three | LeetCode 1262

Medium

Array Dynamic Programming Greedy Sorting

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int maxSumDivThree(int[] nums) {
        int[] dp = {0, 0, 0};
        
        for (int num : nums) {
            int[] dp_copy = dp.clone();
            for (int s : dp_copy) {
                int new_sum = s + num;
                int remainder = new_sum % 3;
                dp[remainder] = Math.max(dp[remainder], new_sum);
            }
        }
        
        return dp[0];
    }
}

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Problem Info

Description

Problem 1262: Greatest Sum Divisible by Three - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Greedy Sorting

Companies

DE Shaw

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1262

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize a `dp` array of size 3 to [0, 0, 0]. `dp[i]` represents the maximum sum with remainder `i` when divided by 3.
2
Step 2: Iterate through each number `num` in the input array `nums`.
3
Step 3: Create a copy of the `dp` array (e.g., `dp_copy`) to avoid modifying it directly during the inner loop. This is crucial because updates to `dp` based on previous values in `dp` are performed.
4
Step 4: Iterate through each element `s` in `dp_copy`.
5
Step 5: Calculate the new sum: `new_sum = s + num`.
6
Step 6: Calculate the remainder when `new_sum` is divided by 3: `remainder = new_sum % 3`.
7
Step 7: Update `dp[remainder]` to the maximum of its current value and `new_sum`: `dp[remainder] = max(dp[remainder], new_sum)`.
8
Step 8: After iterating through all numbers in `nums`, return `dp[0]`, which represents the maximum sum divisible by 3.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because we iterate through the input array `nums` only once. The space complexity is constant because the size of the `dp` array is fixed at 3, regardless of the size of the input array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute-force approach

Generate all possible subsets of the array and calculate the sum of each subset. Check if the sum is divisible by 3 and keep track of the maximum sum.

Time: O(2^n), where n is the length of the input array. This is because there are 2^n possible subsets.Space: O(n), in the worst case, for storing a subset.

Trade-offs: The brute-force approach is simple to understand but very inefficient for large input arrays. It is not practical for the given constraints.

✨

Key Insights

1Insight 1: We only care about the remainders when dividing each number by 3. A number can only have remainders 0, 1, or 2 when divided by 3.
2Insight 2: Dynamic programming can be used to keep track of the maximum sum achievable for each possible remainder (0, 1, 2) when iterating through the array.
3Insight 3: We don't need to store all intermediate sums; we only need to store the largest sum seen so far for each remainder (0, 1, 2).

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input array. The solution handles this implicitly because the `dp` array is initialized to [0, 0, 0], and if the input array is empty, the function will return `dp[0]`, which is 0.
Edge case 2: Input array contains only numbers not divisible by 3. The solution will still return 0, as no combination of numbers will result in a sum divisible by 3.
Edge case 3: Input array contains negative numbers. The provided solutions do not explicitly handle negative numbers, and rely on the problem statement to state that `1 <= nums[i] <= 10^4`. If negative numbers were allowed, the initialization of the `dp` array would have to be adjusted, and a more complex dynamic programming approach involving tracking minimum sums for each remainder would be necessary.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not creating a copy of the `dp` array before iterating through it. If you modify the `dp` array directly during the inner loop, you will be using the updated values in subsequent calculations within the same iteration, leading to incorrect results. The `dp_copy` array is crucial for maintaining the correct state for each number in `nums`.

✗Common mistake 2: Incorrectly initializing the `dp` array. The `dp` array must be initialized to [0, 0, 0] to represent the initial state where no numbers have been selected yet, and therefore the maximum sum for each remainder is 0.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing resource allocation. Imagine you have a set of tasks, each taking a specific amount of time to complete. You want to select a subset of tasks such that the total time is divisible by 3 (e.g., for scheduling purposes).

🏦Real-world scenario 2: Financial portfolio management. You want to select a set of investments such that the total investment amount is divisible by a certain number (e.g., 1000) to simplify accounting.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 3, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Greatest Sum Divisible by Three"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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