# 1304. Find N Unique Integers Sum up to Zero

## Description

Given an integer n, return any array containing n unique integers such that they add up to 0.

Example 1:

Input: n = 5
Output: [-7,-1,1,3,4]
Explanation: These arrays also are accepted [-5,-1,1,2,3] , [-3,-1,2,-2,4].


Example 2:

Input: n = 3
Output: [-1,0,1]


Example 3:

Input: n = 1
Output: [0]


Constraints:

• 1 <= n <= 1000

## Solutions

• class Solution {
public int[] sumZero(int n) {
int[] ans = new int[n];
for (int i = 1, j = 0; i <= n / 2; ++i) {
ans[j++] = i;
ans[j++] = -i;
}
return ans;
}
}

• class Solution {
public:
vector<int> sumZero(int n) {
vector<int> ans(n);
for (int i = 1, j = 0; i <= n / 2; ++i) {
ans[j++] = i;
ans[j++] = -i;
}
return ans;
}
};

• class Solution:
def sumZero(self, n: int) -> List[int]:
ans = []
for i in range(n >> 1):
ans.append(i + 1)
ans.append(-(i + 1))
if n & 1:
ans.append(0)
return ans


• func sumZero(n int) []int {
ans := make([]int, n)
for i, j := 1, 0; i <= n/2; i, j = i+1, j+1 {
ans[j] = i
j++
ans[j] = -i
}
return ans
}

• function sumZero(n: number): number[] {
const ans = new Array(n).fill(0);
for (let i = 1, j = 0; i <= n / 2; ++i) {
ans[j++] = i;
ans[j++] = -i;
}
return ans;
}