Determine the Minimum Sum of a k-avoiding Array

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

Math Greedy Medium

You need to construct a special array with some interesting constraints! Given two integers n and k, your task is to build a k-avoiding array of length n that has the minimum possible sum.

A k-avoiding array is an array of distinct positive integers where no two different elements sum up to k. Think of it as a "forbidden sum" constraint - you must be strategic about which numbers you include!

Goal: Find the minimum possible sum of such an array.

Example: If n = 3 and k = 7, you could create the array [1, 2, 3] since no pair sums to 7, giving a sum of 6.

Input & Output

example_1.py — Basic case

$ Input: n = 5, k = 4

› Output: 18

💡 Note: We can form the array [1, 2, 5, 6, 7]. No two elements sum to 4: 1+2=3, 1+3 would be 4 (but 3 is not in array), etc. Sum = 1+2+5+6+7 = 21. Actually optimal is [1, 2, 5, 6, 8] = 22. Wait, let me recalculate: [1, 2, 5, 6, 7] where we avoid 3 (since 1+3=4) and 4 (since any number + some other = 4). The optimal array is [1, 2, 5, 6, 7] with sum 21. Actually for k=4, we avoid pairs that sum to 4. So we can't have both 1 and 3, can't have both 2 and 2 (but we need distinct), etc. The answer should be [1, 2, 5, 6, 7] = 21, but let me double-check: we want [1, 2, 5, 6, 7], check pairs: 1+2=3✓, 1+5=6✓, 1+6=7✓, 1+7=8✓, 2+5=7✓, 2+6=8✓, 2+7=9✓, 5+6=11✓, 5+7=12✓, 6+7=13✓. Wait, I should be more systematic. For k=4, forbidden pairs are (1,3), (2,2 - invalid since distinct), etc. So we can use 1 but not 3, or use 3 but not 1. To minimize sum, we choose 1. We can use 2. We can't use another 2. We can use 5, 6, 7, etc. So optimal array [1, 2, 5, 6, 7] sum=21. But let me verify this is indeed minimal... Actually, let me recalculate properly: [1, 2, 5, 6, 7], but we could also try [1, 2, 5, 6, 8] = 22, or [1, 2, 5, 7, 8] = 23, etc. The first valid one should be optimal.

example_2.py — Small case

$ Input: n = 3, k = 7

› Output: 6

💡 Note: We can form the array [1, 2, 3]. Let's verify: 1+2=3 (≠7✓), 1+3=4 (≠7✓), 2+3=5 (≠7✓). Sum = 1+2+3 = 6. This is optimal since we're using the three smallest positive integers and none of their pairs sum to 7.

example_3.py — Edge case with k=1

$ Input: n = 2, k = 1

› Output: 3

💡 Note: For k=1, no two positive integers can sum to 1 (since minimum sum of two positive integers is 2). Therefore, any two distinct positive integers form a valid k-avoiding array. The minimum sum is achieved with [1, 2], giving sum = 3.

Constraints

1 ≤ n ≤ 50
1 ≤ k ≤ 100
All array elements must be distinct positive integers
The array must be k-avoiding (no pair sums to k)

Visualization

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Understanding the Visualization

Start Small

Begin with the smallest positive integers to minimize sum

Check Conflicts

For each number, check if its complement (k - number) is already used

Smart Choices

When faced with conflicting pairs like (2,5) for k=7, choose the smaller number

Fill Remaining

Continue until you have n numbers, skipping any that create conflicts

Key Takeaway

🎯 Key Insight: Use greedy construction - always pick the smallest available number that doesn't create a forbidden pair sum. This naturally minimizes the total sum while satisfying the k-avoiding constraint.

Asked in

G Google 15 a Amazon 8 f Meta 6 ⊞ Microsoft 4

The optimal approach uses greedy construction to build the k-avoiding array. We iterate through positive integers starting from 1, and for each number, check if adding it would create a forbidden pair (sum to k) with any existing number. If not, we add it to our result. This ensures we always pick the smallest possible numbers, minimizing the sum. The key insight is that numbers pair up as (1, k-1), (2, k-2), etc., and from each conflicting pair, we should always choose the smaller number to minimize the sum.

Common Approaches

Approach	Time	Space	Notes
✓ Greedy Construction (Optimal)	O(n²)	O(n)	Greedily select smallest numbers while avoiding pairs that sum to k
Brute Force (Generate All Combinations)	O(C(∞,n) × n²)	O(n)	Try all possible combinations of n distinct positive integers and find the valid one with minimum sum

Greedy Construction (Optimal) — Algorithm Steps

Start with an empty result array and current number = 1
For each position, find the smallest number that doesn't form a forbidden sum with any existing number
Add this number to the result and move to the next position
Continue until we have n numbers
Return the sum of all selected numbers

Visualization

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Step-by-Step Walkthrough

Start with 1

Begin checking from the smallest positive integer

Check Conflicts

For each number, check if its complement (k - num) is already used

Add if Safe

If no conflict, add the number to our result

Continue

Move to next number and repeat until we have n numbers

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

long long minimumSum(int n, int k) {
    int* result = (int*)malloc(n * sizeof(int));
    bool* used = (bool*)calloc(10001, sizeof(bool)); // assuming reasonable bounds
    int current = 1;
    int count = 0;
    
    while (count < n) {
        int complement = k - current;
        bool forbidden = (complement > 0 && complement < 10001 && used[complement] && complement != current);
        
        if (!forbidden && (current >= 10001 || !used[current])) {
            result[count] = current;
            if (current < 10001) used[current] = true;
            count++;
        }
        
        current++;
    }
    
    long long sum = 0;
    for (int i = 0; i < n; i++) {
        sum += result[i];
    }
    
    free(result);
    free(used);
    return sum;
}

int main() {
    int n, k;
    scanf("%d %d", &n, &k);
    printf("%lld\n", minimumSum(n, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n positions, we might need to check up to n existing numbers for conflicts

⚠ Quadratic Growth

Space Complexity

O(n)

Space to store the result array and track used numbers

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Greedy Construction (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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