I checked from website, this is called unimodal permutation, which defines as a sequence that has only one local maximum. For example n = 5:

```
12345
12354
12453
12543
13452
13542
14532
15432
23451
23541
24531
25431
34521
35421
45321
54321
```

Is there an algorithm to get the kth unimodal permutations?

Answer:

The trick here is to index the sequence in binary

```
1234
----
12345 (0000)
12354 (0001)
12453 (0010)
12543 (0011)
13452 (0100)
13542 (0101)
14532 (0110)
15432 (0111)
23451 (1000)
23541 (1001)
24531 (1010)
25431 (1011)
34521 (1100)
35421 (1101)
45321 (1110)
54321 (1111)
```

and then observe that the numbers 1..4 appear before the 5 if and only if their bit is 0. In Python:

```
def kth_unimodal(n, k): # k is 0-indexed
left = []
right = []
for j in range(1, n): # 1..n-1
if k & (1 << (n - 1 - j)):
right.append(j)
else:
left.append(j)
left.append(n)
left.extend(reversed(right))
return left
```

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