Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.
Example:
Given n = 2, return 91. (The answer should be the total numbers in the range of 0 ≤ x < 100, excluding [11,22,33,44,55,66,77,88,99])
Credits:
Special thanks to @memoryless for adding this problem and creating all test cases.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 | class Solution: def countNumbersWithUniqueDigits(self, n): """ :type n: int :rtype: int """ used = [False for x in range(10)] def gen(used, n, d): if n == d: return 1 total = 1 startIndex = 1 if d == 0 else 0 for i in range(startIndex, 10): if not used[i]: used[i] = True total += gen(used, n, d + 1) used[i] = False return total return gen(used, n, 0)s = Solution()a = s.countNumbersWithUniqueDigits(7)print(a) |
class Solution:
def countNumbersWithUniqueDigits(self, n):
"""
:type n: int
:rtype: int
"""
used = [False for x in range(10)]
def gen(used, n, d):
if n == d:
return 1
total = 1
startIndex = 1 if d == 0 else 0
for i in range(startIndex, 10):
if not used[i]:
used[i] = True
total += gen(used, n, d + 1)
used[i] = False
return total
return gen(used, n, 0)
s = Solution()
a = s.countNumbersWithUniqueDigits(7)
print(a)