[
strign
greedy
]
BinarySearch 0374 Consecutively Descending Integers
Problem statement
https://binarysearch.com/problems/Consecutively-Descending-Integers/
Solution
Choose as starting number s[:1], s[:2], ...
and so on.
Complexity
It is O(n^2)
for time and O(n)
for space.
Code
class Solution:
def solve(self, s):
def helper(s, n):
if not s: return True
if n < 0: return False
sn = str(n)
if s[:len(sn)] == sn:
return helper(s[len(sn):], n - 1)
return False
for l in range(1, len(s)):
n = int(s[:l])
if helper(s[l:], n - 1):
return True
return False