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

Starting from the positive integer $i = 1$, we sequentially determine if $i$ can be added to the array. 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. We continue this process until the array's length reaches $n$.

The time complexity is $O(n + k)$, and the space complexity is $O(n + k)$. Where $n$ is the length of the array.

Python3

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
            vis.add(k - i)
            s += i
            i += 1
        return s

Java

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

C++

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

Go

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

TypeScript

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

Rust

impl Solution {
    pub fn minimum_sum(n: i32, k: i32) -> i32 {
        let (mut s, mut i) = (0, 1);
        let mut vis = std::collections::HashSet::new();

        for _ in 0..n {
            while vis.contains(&i) {
                i += 1;
            }
            vis.insert(k - i);
            s += i;
            i += 1;
        }

        s
    }
}