"""
Given an integer n, return 1 - n in lexicographical order.

For example, given 13, return: [1,10,11,12,13,2,3,4,5,6,7,8,9].

Please optimize your algorithm to use less time and space. The input size may be
as large as 5,000,000.
"""
class Solution(object):
    def lexicalOrder(self, n):
        """
        :type n: int
        :rtype: List[int]
        """
        soln = [None] * n
        curr = 1
        for i in range(n):
            soln[i] = curr
            if curr * 10 <= n:
                curr *= 10
            else:
                if curr >= n:
                    curr //= 10
                curr += 1
                while curr % 10 == 0:
                    curr //= 10
        return soln


a = Solution()
print(a.lexicalOrder(13) == [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9])

"""
Note: easy but expensive solution, not accepted
    def lexicalOrder(self, n):
        return sorted([i + 1 for i in range(n)], key=str)
"""

"""
Given an integer n, return 1 - n in lexicographical order.

For example, given 13, return: [1,10,11,12,13,2,3,4,5,6,7,8,9].

Please optimize your algorithm to use less time and space. The input size may be
as large as 5,000,000.
"""
class Solution(object):
    def lexicalOrder(self, n):
        """
        :type n: int
        :rtype: List[int]
        """
        soln = [None] * n
        curr = 1
        for i in range(n):
            soln[i] = curr
            if curr * 10 <= n:
                curr *= 10
            else:
                if curr >= n:
                    curr //= 10
                curr += 1
                while curr % 10 == 0:
                    curr //= 10
        return soln


a = Solution()
print(a.lexicalOrder(13) == [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9])

"""
Note: easy but expensive solution, not accepted
    def lexicalOrder(self, n):
        return sorted([i + 1 for i in range(n)], key=str)
"""

Given an integer n, return 1 - n in lexicographical order.  For example, given 13, return: [1,10,11,12,13,2,3,4,5,6,7,8,9].  Please optimize your algorithm to use less time and space. The input size may be as large as 5,000,000. :type n: int         :rtype: List[int] Note: easy but expensive solution, not accepted     def lexicalOrder(self, n):         return sorted([i + 1 for i in range(n)], key=str)