Given an integer n, return the nth digit of the infinite integer sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...].
Example 1:
Input: n = 3
Output: 3
Example 2:
Input: n = 11
Output: 0
Explanation: The 11th digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... is a 0, which is part of the number 10.
Constraints:
-
1 <= n <= 231 - 1
SOLUTION:
class Solution:
def findNthDigit(self, n: int) -> int:
prevcumsum = None
cumsum = 0
l = 1
while n >= cumsum:
prevcumsum = cumsum
nextcumsum = cumsum + 9 * l * (10 ** (l - 1))
if nextcumsum >= n:
break
cumsum = nextcumsum
l += 1
n -= prevcumsum
k, n = divmod(n - 1, l)
num = str(10 ** (l - 1) + k)
return int(num[n])
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