What is the time complexity of Determine the Minimum Sum of a k-avoiding Array?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Determine the Minimum Sum of a k-avoiding Array?

This problem uses Math, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Determine the Minimum Sum of a k-avoiding Array?

Insight 1: The smallest possible numbers to include in the array are 1, 2, 3, and so on. We want to use as many of these as possible while still adhering to the 'k-avoiding' constraint. Insight 2: If we include a number 'x' in the array, we cannot include 'k - x'. This defines the primary constraint we must satisfy to maintain the k-avoiding property.

Which companies ask LeetCode 2829. Determine the Minimum Sum of a k-avoiding Array in interviews?

This problem is frequently asked at Infosys.

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Determine the Minimum Sum of a k-avoiding Array - LeetCode 2829 Solution

Determine the Minimum Sum of a k-avoiding Array - Complete Solution Guide

Determine the Minimum Sum of a k-avoiding Array is LeetCode problem 2829, 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

You are given two integers, n and k . An array of distinct positive integers is called a k-avoiding array if there does not exist any pair of distinct elements that sum to k . Return the minimum possible sum of a k-avoiding array of length n . Example 1: Input: n = 5, k = 4 Output: 18 Explanation: Consider the k-avoiding array [1,2,4,5,6], which has a sum of 18. It can be proven that there is no k-avoiding array with a sum less than 18. Example 2: Input: n = 2, k = 6 Output: 3 Explanation: We ca

Detailed Explanation

The problem requires finding the minimum possible sum of an array of 'n' distinct positive integers, with the constraint that no two distinct elements in the array sum up to 'k'. In other words, we need to construct a 'k-avoiding' array of length 'n' with the smallest possible sum.

Solution Approach

The solution is based on the idea that we should start by including the smallest positive integers (1, 2, 3, ...) in our k-avoiding array. We calculate 'm' as k/2 (integer division). If n <= m, it means we can take the first n positive integers without violating the k-avoiding property. If n > m, we include the numbers from 1 to m, and then include the next n - m numbers starting from k. This strategy guarantees the minimum possible sum because we take the smallest numbers until we can no longer do so without violating the 'k-avoiding' constraint, then shift to numbers greater than or equal to 'k'.

Step-by-Step Algorithm

Step 1: Calculate m = k // 2 (integer division) or k / 2 (depending on the language).
Step 2: Check if n <= m. If it is, the sum is the sum of the first n positive integers: n * (n + 1) // 2.
Step 3: If n > m, calculate the sum of the first m positive integers: m * (m + 1) // 2. This is the sum of the first part of the array (numbers less than k/2).
Step 4: Calculate the remaining count of elements to include: remaining_count = n - m.
Step 5: The next elements we add start at k, k+1, k+2... so we calculate the sum of these remaining elements: sum_second_part = remaining_count * k + (remaining_count * (remaining_count - 1)) // 2. This can also be thought of as the sum of an arithmetic sequence.
Step 6: Return the sum of the first part and the second part: sum_first_part + sum_second_part.

Key Insights

Insight 1: The smallest possible numbers to include in the array are 1, 2, 3, and so on. We want to use as many of these as possible while still adhering to the 'k-avoiding' constraint.
Insight 2: If we include a number 'x' in the array, we cannot include 'k - x'. This defines the primary constraint we must satisfy to maintain the k-avoiding property.
Insight 3: It's optimal to include numbers less than k/2 if we can, and then include numbers greater than or equal to k to fill the array up to size n.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Math, Greedy.

Companies

Asked at: Infosys.

Determine the Minimum Sum of a k-avoiding Array | LeetCode 2829

Medium

Math Greedy

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

Solution Code

Available in:

class Solution {
    public int minimumSum(int n, int k) {
        int m = k / 2;
        
        if (n <= m) {
            return (n * (n + 1)) / 2;
        } else {
            int sum_first_part = (m * (m + 1)) / 2;
            int remaining_count = n - m;
            int sum_second_part = remaining_count * k + (remaining_count * (remaining_count - 1)) / 2;
            return sum_first_part + sum_second_part;
        }
    }
}

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Problem Info

Description

Problem 2829: Determine the Minimum Sum of a k-avoiding Array - Detailed explanation and solution approaches.

Topics

Math Greedy

Companies

Infosys

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2829

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

6 steps

1
Step 1: Calculate m = k // 2 (integer division) or k / 2 (depending on the language).
2
Step 2: Check if n <= m. If it is, the sum is the sum of the first n positive integers: n * (n + 1) // 2.
3
Step 3: If n > m, calculate the sum of the first m positive integers: m * (m + 1) // 2. This is the sum of the first part of the array (numbers less than k/2).
4
Step 4: Calculate the remaining count of elements to include: remaining_count = n - m.
5
Step 5: The next elements we add start at k, k+1, k+2... so we calculate the sum of these remaining elements: sum_second_part = remaining_count * k + (remaining_count * (remaining_count - 1)) // 2. This can also be thought of as the sum of an arithmetic sequence.
6
Step 6: Return the sum of the first part and the second part: sum_first_part + sum_second_part.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The solution involves a few arithmetic operations (division, multiplication, addition). These operations take constant time regardless of the input values of 'n' and 'k'. No additional data structures that depend on the input size are created, so space usage is constant.

📖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

Iterative Construction

An alternative approach is to build the k-avoiding array iteratively. Start with an empty array and add numbers from 1 upwards. Before adding a number 'x', check if 'k - x' is already present in the array. If it is, skip 'x' and try the next number. Continue this process until the array has 'n' elements.

Time: O(n*m) in the worst case, where m is the largest number in the constructed array. This is because we may need to check for many potential numbers and iterate through the existing array each time.Space: O(n) to store the k-avoiding array.

Trade-offs: This approach is less efficient than the O(1) solution. It might be easier to understand initially, but it's not suitable for larger values of 'n' and 'k'.

✨

Key Insights

1Insight 1: The smallest possible numbers to include in the array are 1, 2, 3, and so on. We want to use as many of these as possible while still adhering to the 'k-avoiding' constraint.
2Insight 2: If we include a number 'x' in the array, we cannot include 'k - x'. This defines the primary constraint we must satisfy to maintain the k-avoiding property.
3Insight 3: It's optimal to include numbers less than k/2 if we can, and then include numbers greater than or equal to k to fill the array up to size n.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: n = 1. The solution works correctly, as it will simply return the sum of the first element, which will always be 1 unless k=2 (1 is k-avoiding). If k = 2 and n = 1 the algorithm will return 1 as the minimum sum.
Edge case 2: n = k. The algorithm will calculate the numbers [1...k/2] and then [k, k+1, ...]. This maintains the k-avoiding property.
Edge case 3: k is a small number. For example, if k=2, the solution correctly identifies that you cannot include 1 with any other number, so it fills up the n spots with 2, 3, 4.. etc.
Edge case 4: k is greater than n. This means that all numbers from 1 to n can be included as there will be no two elements adding up to k.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to use integer division when calculating 'm = k // 2'. Using regular division can lead to incorrect results.

✗Common mistake 2: Not considering the case where 'n' is smaller than or equal to 'm'. In this case, the sum is simply the sum of the first 'n' positive integers.

✗Common mistake 3: Failing to add the remaining numbers starting from 'k'. Instead, some beginners might continue the sequence from 'm+1' leading to including pairs that sums up to k.

✗Common mistake 4: Incorrectly calculating the sum of an arithmetic series.

✅

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: Resource allocation. Imagine you need to allocate 'n' resources with constraints. The 'k-avoiding' concept could represent a constraint where certain pairs of resources cannot be used together due to incompatibility.

🏦Real-world scenario 2: Cryptography. Selecting a set of numbers for cryptographic keys where certain combinations could compromise security. The k-avoiding property could be used to avoid generating vulnerable key combinations.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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: October 8, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Determine the Minimum Sum of a k-avoiding Array"?

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

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