# 667. Beautiful Arrangement II

## Description

Given two integers n and k, construct a list answer that contains n different positive integers ranging from 1 to n and obeys the following requirement:

• Suppose this list is answer = [a1, a2, a3, ... , an], then the list [|a1 - a2|, |a2 - a3|, |a3 - a4|, ... , |an-1 - an|] has exactly k distinct integers.

Return the list answer. If there multiple valid answers, return any of them.

Example 1:

Input: n = 3, k = 1
Output: [1,2,3]
Explanation: The [1,2,3] has three different positive integers ranging from 1 to 3, and the [1,1] has exactly 1 distinct integer: 1


Example 2:

Input: n = 3, k = 2
Output: [1,3,2]
Explanation: The [1,3,2] has three different positive integers ranging from 1 to 3, and the [2,1] has exactly 2 distinct integers: 1 and 2.


Constraints:

• 1 <= k < n <= 104

## Solutions

• class Solution {
public int[] constructArray(int n, int k) {
int l = 1, r = n;
int[] ans = new int[n];
for (int i = 0; i < k; ++i) {
ans[i] = i % 2 == 0 ? l++ : r--;
}
for (int i = k; i < n; ++i) {
ans[i] = k % 2 == 0 ? r-- : l++;
}
return ans;
}
}

• class Solution {
public:
vector<int> constructArray(int n, int k) {
int l = 1, r = n;
vector<int> ans(n);
for (int i = 0; i < k; ++i) {
ans[i] = i % 2 == 0 ? l++ : r--;
}
for (int i = k; i < n; ++i) {
ans[i] = k % 2 == 0 ? r-- : l++;
}
return ans;
}
};

• class Solution:
def constructArray(self, n: int, k: int) -> List[int]:
l, r = 1, n
ans = []
for i in range(k):
if i % 2 == 0:
ans.append(l)
l += 1
else:
ans.append(r)
r -= 1
for i in range(k, n):
if k % 2 == 0:
ans.append(r)
r -= 1
else:
ans.append(l)
l += 1
return ans


• func constructArray(n int, k int) []int {
l, r := 1, n
ans := make([]int, n)
for i := 0; i < k; i++ {
if i%2 == 0 {
ans[i] = l
l++
} else {
ans[i] = r
r--
}
}
for i := k; i < n; i++ {
if k%2 == 0 {
ans[i] = r
r--
} else {
ans[i] = l
l++
}
}
return ans
}

• function constructArray(n: number, k: number): number[] {
let l = 1;
let r = n;
const ans = new Array(n);
for (let i = 0; i < k; ++i) {
ans[i] = i % 2 == 0 ? l++ : r--;
}
for (let i = k; i < n; ++i) {
ans[i] = k % 2 == 0 ? r-- : l++;
}
return ans;
}