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# Program to construct the lexicographically largest valid sequence in Python

Suppose we have a number n, we have to find a sequence that satisfies all of the following rules −

1 occurs once in the sequence.

Each number in between 2 and n occurs twice in the sequence.

For every i in range 2 to n, the distance between the two occurrences of i is exactly i.

The distance between two numbers on the sequence, a[i] and a[j], is |j - i|. We have to find the lexicographically largest sequence.

So, if the input is like n = 4, then the output will be [4, 2, 3, 2, 4, 3, 1]

To solve this, we will follow these steps −

- Define a function backtrack() . This will take elems, res, start := 0
- if size of elems <= 0, then
- return True

- if start >= size of res , then
- return False

- if res[start] is not same as -1, then
- return backtrack(elems, res, start + 1)

- for i in range 0 to size of elems - 1, do
- num := elems[i]
- dist := 0 when num is same as 1 otherwise num
- if (start + dist) < size of res and res[start + dist] is same as -1, then
- res[start] := num
- res[start + dist] := num
- delete ith element from elems
- if backtrack(elems, res, start) is false, then
- res[start] := -1
- res[start + dist] := -1
- insert num into elems at position i
- go for next iteration

- otherwise,
- return True

- From the main method, do the following
- elems := a list with n elements from n down to 1
- res := a list of size n*2-1 and fill with -1
- backtrack(elems, res)
- return res

## Example

Let us see the following implementation to get better understanding −

def backtrack(elems, res, start = 0): if len(elems) <= 0: return True if start >= len(res): return False if res[start] != -1: return backtrack(elems, res, start + 1) for i in range(len(elems)): num = elems[i] dist = 0 if num == 1 else num if (start + dist) < len(res) and res[start + dist] == -1: res[start] = num res[start + dist] = num elems.pop(i) if not backtrack(elems, res, start): res[start] = -1 res[start + dist] = -1 elems.insert(i, num) continue else: return True def solve(n): elems = [ i for i in range(n,0,-1)] res = [ -1 for i in range(n*2 - 1)] backtrack(elems, res) return res n = 4 print(solve(n))

## Input

4

## Output

[4, 2, 3, 2, 4, 3, 1]

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