Let `1..k`

be the `k`

unique differences.

Let this be the arrangement of differences:

`[k k-1 k-2 k-3 ... 1 1 1 1 1 1 1 1]`

Let the solution start at `[1]`

The solution for for getting these differences would then be:

`[1 k+1 2 k 3 k-1 4 k-2 .. k+2 k+3 .. n]`

**Example**

```
n=10, k=6
1 7 2 6 3 5 4 8 9 10
k 6 -5 4 -3 2 -1 0 0 0 0
```

**Solution**

```
class Solution(object):
def constructArray(self, n, k):
a = [1]
h = k+1
l = 2
for _k in range(k,0,-2):
if _k == 1:
a.append(h)
break
else:
a.append(h)
a.append(l)
h -= 1
l += 1
a.extend(range(k+2,n+1))
return a
```