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

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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

✓ Optimized

⏱️ Time: N/A Space: N/A

Generate All Combinations

⏱️ Time: O(C(∞,n) × n²) Space: O(n)

Generate all possible sets of n distinct positive integers and check each set to see if it's k-avoiding. Return the minimum sum among valid sets.

Greedy Construction

⏱️ Time: O(n²) Space: O(n)

Start from 1 and keep adding the smallest number that doesn't create a pair summing to k. This greedy approach works because we always want the smallest possible sum.

Mathematical Formula

⏱️ Time: O(1) Space: O(1)

Analyze the pattern: when k is small relative to n, we need to skip certain numbers. We can derive a formula based on how many numbers we need to skip.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

int min_global = INT_MAX;

void generate_combinations(int* nums, int size, int n, int k, int start, int* current, int current_size) {
    if (current_size == n) {
        // Check if this combination is k-avoiding
        int is_avoiding = 1;
        for (int i = 0; i < n && is_avoiding; i++) {
            for (int j = i + 1; j < n; j++) {
                if (current[i] + current[j] == k) {
                    is_avoiding = 0;
                    break;
                }
            }
        }
        
        if (is_avoiding) {
            int sum = 0;
            for (int i = 0; i < n; i++) {
                sum += current[i];
            }
            if (sum < min_global) {
                min_global = sum;
            }
        }
        return;
    }
    
    for (int i = start; i < size; i++) {
        current[current_size] = nums[i];
        generate_combinations(nums, size, n, k, i + 1, current, current_size + 1);
    }
}

int solution(int n, int k) {
    min_global = INT_MAX;
    
    // Try combinations starting from smallest possible numbers
    for (int upper_bound = n; upper_bound <= n * k; upper_bound++) {
        int* nums = (int*)malloc(upper_bound * sizeof(int));
        for (int i = 0; i < upper_bound; i++) {
            nums[i] = i + 1;
        }
        
        int* current = (int*)malloc(n * sizeof(int));
        generate_combinations(nums, upper_bound, n, k, 0, current, 0);
        
        free(nums);
        free(current);
        
        if (min_global != INT_MAX) {
            break;
        }
    }
    
    return min_global;
}

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

⏱️

⚠ Quadratic Growth

Space Complexity

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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