Determine the Minimum Sum of a k-avoiding Array

Determine the Minimum Sum of a k-avoiding Array - Problem

Math Greedy Medium

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.

Input & Output

Example 1 — Basic Case

$ Input: n = 3, k = 4

› Output: 8

💡 Note: The k-avoiding array [1,2,5] has the minimum sum. We skip 3 and 4 because 1+3=4 and 2+2=4 (but we need distinct elements, so we avoid the pair (1,3)).

Example 2 — Large k

$ Input: n = 2, k = 6

› Output: 3

💡 Note: Since k=6 is large, we can take the first n=2 numbers: [1,2]. Their sum is 1+2=3, and 1+2≠6 so it's k-avoiding.

Example 3 — Small k

$ Input: n = 2, k = 3

› Output: 4

💡 Note: We cannot use [1,2] because 1+2=3=k. So we take [1,3] with sum 4, avoiding the forbidden pair.

Constraints

1 ≤ n, k ≤ 50

Visualization

Tap to expand

Understanding the Visualization

Input

Given n=3, k=4 - need 3 distinct positive integers

Constraint

No two elements should sum to k=4

Output

Minimum possible sum of such array

Key Takeaway

🎯 Key Insight: Greedily select smallest numbers while checking for forbidden pairs

Asked in

G Google 15 f Meta 12 a Amazon 8

The key insight is to greedily select the smallest positive integers while avoiding pairs that sum to k. The optimal approach uses a greedy strategy with set-based conflict checking. Time: O(n), Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Formula	O(1)	O(1)	Use mathematical insight to directly calculate the minimum sum
Greedy Construction	O(n²)	O(n)	Build the array by greedily selecting the smallest valid numbers
Generate All Combinations	O(C(∞,n) × n²)	O(n)	Try all possible combinations of n distinct positive integers

Mathematical Formula — Algorithm Steps

If k <= n+1, we take first n numbers and skip those that would pair to k
If k > n+1, we can take first n numbers directly
Calculate sum using arithmetic progression formula

Visualization

Tap to expand

Step-by-Step Walkthrough

Analyze

When k > 2n, no conflicts possible with first n numbers

Pattern

For smaller k, skip numbers that pair with previous numbers

Formula

Direct calculation based on mathematical relationship

Code -

solution.c — C

#include <stdio.h>

int solution(int n, int k) {
    // If k is large enough, we can just take the first n positive integers
    if (k > 2 * n) {
        return n * (n + 1) / 2;
    }
    
    // Otherwise, we need to be careful about pairs that sum to k
    int totalSum = 0;
    int count = 0;
    int current = 1;
    
    while (count < n) {
        int pair = k - current;
        if (pair > current) {  // current is smaller, take it
            totalSum += current;
            count++;
        } else if (pair < current) {  // pair is smaller and already processed
            totalSum += current;
            count++;
        }
        // else pair == current, skip this number
        
        current++;
    }
    
    return totalSum;
}

int main() {
    int n, k;
    scanf("%d", &n);
    scanf("%d", &k);
    int result = solution(n, k);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Direct mathematical calculation without loops

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

3.2K Views

Medium Frequency

~15 min Avg. Time

145 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Mathematical Formula — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler