Sequential Digits in Python

# Sequential Digits in Python

## Problem:

An integer has sequential digits if and only if each digit in the number is one more than the previous digit.

Return a sorted list of all the integers in the range `[low, high]` inclusive that have sequential digits.

## Example:

``````Input: low = 100, high = 300

Output: [123,234]``````

## Constraints:

``10 <= low <= high <= 10^9``

## Solution:

``````class Solution:
def sequentialDigits(low, high):
result = []
start = int(str(low))
for val in range(1, len(str(low))):
new_val = start%10 + 1
start = start*10 + new_val
if start > high:
return result

result.append(start)

while result[-1] <= high:
temp = str(result[-1])
next_elem = int(temp[-1]) + 1

if next_elem > 9:
next_greater = 0
for index in range(len(temp) + 1):
next_greater = next_greater*10 + (index+1)
else:
next_greater = int(temp[1:]) * 10 + next_elem

if next_greater <= high:
result.append(next_greater)
else:
break

final_result = []
for val in result:
if '0' not in str(val) and val >= low:
final_result.append(val)
return final_result

low = 100
high = 300
print(Solution.sequentialDigits(100, 300))``````