What is the time complexity of Find the Minimum Possible Sum of a Beautiful Array?

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

What data structures are used to solve Find the Minimum Possible Sum of a Beautiful Array?

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

What are the key insights for solving Find the Minimum Possible Sum of a Beautiful Array?

Insight 1: To minimize the sum, we should start with the smallest positive integers (1, 2, 3, ...). Insight 2: We can include numbers up to target/2 without violating the 'beautiful' condition because if we pick a number 'x' less than or equal to target/2, there is no other distinct 'y' less than or equal to target/2 that can be added to 'x' to get 'target'. Numbers greater than or equal to 'target' do not cause conflicts, but we want to use as few of those as possible, so we use the first 'n' beautiful numbers.

Which companies ask LeetCode 2834. Find the Minimum Possible Sum of a Beautiful Array in interviews?

This problem is frequently asked at Infosys.

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Find the Minimum Possible Sum of a Beautiful Array - LeetCode 2834 Solution

Find the Minimum Possible Sum of a Beautiful Array - Complete Solution Guide

Find the Minimum Possible Sum of a Beautiful Array is LeetCode problem 2834, 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 positive integers n and target . An array nums is beautiful if it meets the following conditions: nums.length == n . nums consists of pairwise distinct positive integers. There doesn't exist two distinct indices, i and j , in the range [0, n - 1] , such that nums[i] + nums[j] == target . Return the minimum possible sum that a beautiful array could have modulo 10 9 + 7 . Example 1: Input: n = 2, target = 3 Output: 4 Explanation: We can see that nums = [1,3] is beautiful. - The array

Detailed Explanation

The problem asks us to find the minimum possible sum of a "beautiful" array. An array is beautiful if it has 'n' distinct positive integers and no two distinct elements in the array sum up to 'target'. The goal is to find an array that satisfies these conditions and has the smallest possible sum, then return this sum modulo 10^9 + 7.

Solution Approach

The solution uses a greedy approach. It first calculates how many numbers less than or equal to `target / 2` can be used. If `n` is less than or equal to `target / 2`, it uses numbers from 1 to `n`. Otherwise, it uses numbers from 1 to `target / 2`, and the remaining numbers are taken starting from `target`. The sum of each part is calculated separately using the formula for the sum of an arithmetic series (optimized by splitting into term1 and term2 for handling modulo without overflow), and finally, these sums are added, taking the modulo at each stage to prevent integer overflow.

Step-by-Step Algorithm

Step 1: Calculate `m = target / 2`. This represents the number of integers we can use without conflicting with the target sum rule from 1 to m.
Step 2: If `n <= m`, then the array can be constructed using the first `n` positive integers. Calculate the sum of numbers from 1 to `n` using the `sum_1_to_x` function and return the result modulo 10^9 + 7.
Step 3: If `n > m`, then we need to use integers greater than or equal to `target`. Calculate the sum of numbers from 1 to `m` (sum1).
Step 4: Calculate `k = n - m`. This is the number of integers needed from the range starting at `target`.
Step 5: Calculate the starting value for the numbers greater than `target/2` (`first_val_part2 = target`). Calculate the last value for this second part `last_val_part2 = target + k - 1`.
Step 6: Calculate the sum of numbers from `first_val_part2` to `last_val_part2` by finding the sum of numbers from 1 to `last_val_part2` and subtracting the sum of numbers from 1 to `first_val_part2 - 1` (sum2).
Step 7: Add `sum1` and `sum2`, take the modulo, and return the result.

Key Insights

Insight 1: To minimize the sum, we should start with the smallest positive integers (1, 2, 3, ...).
Insight 2: We can include numbers up to target/2 without violating the 'beautiful' condition because if we pick a number 'x' less than or equal to target/2, there is no other distinct 'y' less than or equal to target/2 that can be added to 'x' to get 'target'. Numbers greater than or equal to 'target' do not cause conflicts, but we want to use as few of those as possible, so we use the first 'n' beautiful numbers.
Insight 3: The problem can be divided into two cases. Either we can construct a beautiful array of size 'n' using numbers less than 'target', or we need to use numbers greater than or equal to 'target' after we've exhausted using the numbers less than target/2. We use the formula for the sum of an arithmetic series to calculate the sums efficiently and handle the modulo operation carefully.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Math, Greedy.

Companies

Asked at: Infosys.

Find the Minimum Possible Sum of a Beautiful Array | LeetCode 2834

Medium

Math Greedy

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

Solution Code

Available in:

class Solution {
    public int minimumPossibleSum(int n, int target) {
        long MOD = 1000000007;

        
        
        
        
        
        
        

        
        
        
        
        
        
        
        
        
        
        
        

        long m = target / 2;

        if (n <= m) {
            return (int)sum_1_to_x(n, MOD);
        } else {
            long sum1 = sum_1_to_x(m, MOD);
            
            long k = n - m;
            
            long first_val_part2 = target;
            long last_val_part2 = target + k - 1;
            
            long sum_up_to_last = sum_1_to_x(last_val_part2, MOD);
            long sum_up_to_first_minus_1 = sum_1_to_x(first_val_part2 - 1, MOD);
            
            long sum2 = (sum_up_to_last - sum_up_to_first_minus_1 + MOD) % MOD;
            
            long total_sum = (sum1 + sum2) % MOD;
            return (int)total_sum;
        }
    }

    private long sum_1_to_x(long x, long MOD) {
        if (x <= 0) {
            return 0;
        }
        
        long term1;
        long term2;
        if (x % 2 == 0) {
            term1 = (x / 2) % MOD;
            term2 = (x + 1) % MOD;
        } else {
            term1 = x % MOD;
            term2 = ((x + 1) / 2) % MOD;
        }
        
        return (term1 * term2) % MOD;
    }
}

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Problem Info

Description

Problem 2834: Find the Minimum Possible Sum of a Beautiful Array - Detailed explanation and solution approaches.

Topics

Math Greedy

Companies

Infosys

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2834

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

7 steps

1
Step 1: Calculate `m = target / 2`. This represents the number of integers we can use without conflicting with the target sum rule from 1 to m.
2
Step 2: If `n <= m`, then the array can be constructed using the first `n` positive integers. Calculate the sum of numbers from 1 to `n` using the `sum_1_to_x` function and return the result modulo 10^9 + 7.
3
Step 3: If `n > m`, then we need to use integers greater than or equal to `target`. Calculate the sum of numbers from 1 to `m` (sum1).
4
Step 4: Calculate `k = n - m`. This is the number of integers needed from the range starting at `target`.
5
Step 5: Calculate the starting value for the numbers greater than `target/2` (`first_val_part2 = target`). Calculate the last value for this second part `last_val_part2 = target + k - 1`.
6
Step 6: Calculate the sum of numbers from `first_val_part2` to `last_val_part2` by finding the sum of numbers from 1 to `last_val_part2` and subtracting the sum of numbers from 1 to `first_val_part2 - 1` (sum2).
7
Step 7: Add `sum1` and `sum2`, take the modulo, and return the result.

💡

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 performs a fixed number of arithmetic operations. The `sum_1_to_x` function is called a fixed number of times, and it only performs a constant number of calculations. Therefore, the time complexity is O(1). The solution only uses a constant amount of extra space to store variables, independent of the input size. Therefore, the space complexity is O(1).

📖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 approach

Iteratively construct the array and sum the elements while ensuring distinctness and the 'beautiful' condition. This would likely involve using a Set to track numbers that would cause issues if added, due to the target sum property.

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

Trade-offs: The iterative approach is less efficient than the current approach (O(1)) because it has a O(n) time complexity. It also needs O(n) space to maintain the set. Using the iterative approach could potentially lead to Time Limit Exceeded (TLE) errors on large datasets.

✨

Key Insights

1Insight 1: To minimize the sum, we should start with the smallest positive integers (1, 2, 3, ...).
2Insight 2: We can include numbers up to target/2 without violating the 'beautiful' condition because if we pick a number 'x' less than or equal to target/2, there is no other distinct 'y' less than or equal to target/2 that can be added to 'x' to get 'target'. Numbers greater than or equal to 'target' do not cause conflicts, but we want to use as few of those as possible, so we use the first 'n' beautiful numbers.
3Insight 3: The problem can be divided into two cases. Either we can construct a beautiful array of size 'n' using numbers less than 'target', or we need to use numbers greater than or equal to 'target' after we've exhausted using the numbers less than target/2. We use the formula for the sum of an arithmetic series to calculate the sums efficiently and handle the modulo operation carefully.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: `n = 1`, `target = 1`. The solution will correctly return 1 because sum_1_to_x(1) = 1.
Edge case 2: Large values of `n` and `target` where integer overflow might occur. The solution uses modulo operations `% MOD` at each step to prevent integer overflow.
Edge case 3: `target = 1`. Since `m = 0`, then all `n` numbers used will be greater than or equal to 1. Thus the beautiful numbers are `1, 2, 3,... n`. The resulting sum is the summation of the integers from 1 to n.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow when calculating the sum without using modulo operator.

✗Common mistake 2: Not handling the case where `n > target / 2` correctly. It is important to take the next available integer from the beginning after having exhausted the first `target / 2` integers.

✗Common mistake 3: Forgetting the modulo operation when returning the final result.

✅

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 problems where you need to minimize cost while satisfying specific constraints (similar to the 'beautiful' array condition).

🏦Real-world scenario 2: Cryptographic key generation. You might want to choose a set of keys with minimal sum, subject to certain conditions to prevent attacks.

💡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

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 13, 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 13

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

@DevExpert92Jul 6

❓

Frequently Asked Questions

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

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 "Find the Minimum Possible Sum of a Beautiful 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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