# 2829. Determine the Minimum Sum of a k-avoiding Array

## Description

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 can construct the array [1,2], which has a sum of 3.
It can be proven that there is no k-avoiding array with a sum less than 3.


Constraints:

• 1 <= n, k <= 50

## Solutions

Solution 1: Greedy + Simulation

We start from the positive integer $i=1$, and judge whether $i$ can be added to the array in turn. If it can be added, we add $i$ to the array, accumulate it to the answer, and then mark $k-i$ as visited, indicating that $k-i$ cannot be added to the array. The loop continues until the length of the array is $n$.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of the array.

• class Solution {
public int minimumSum(int n, int k) {
int s = 0, i = 1;
boolean[] vis = new boolean[k + n * n + 1];
while (n-- > 0) {
while (vis[i]) {
++i;
}
vis[i] = true;
if (k >= i) {
vis[k - i] = true;
}
s += i;
}
return s;
}
}

• class Solution {
public:
int minimumSum(int n, int k) {
int s = 0, i = 1;
bool vis[k + n * n + 1];
memset(vis, false, sizeof(vis));
while (n--) {
while (vis[i]) {
++i;
}
vis[i] = true;
if (k >= i) {
vis[k - i] = true;
}
s += i;
}
return s;
}
};

• class Solution:
def minimumSum(self, n: int, k: int) -> int:
s, i = 0, 1
vis = set()
for _ in range(n):
while i in vis:
i += 1
s += i
return s


• func minimumSum(n int, k int) int {
s, i := 0, 1
vis := make([]bool, k+n*n+1)
for ; n > 0; n-- {
for vis[i] {
i++
}
vis[i] = true
if k >= i {
vis[k-i] = true
}
s += i
}
return s
}

• function minimumSum(n: number, k: number): number {
let s = 0;
let i = 1;
const vis: boolean[] = Array(n * n + k + 1);
while (n--) {
while (vis[i]) {
++i;
}
vis[i] = true;
if (k >= i) {
vis[k - i] = true;
}
s += i;
}
return s;
}