class Solution:
def lastSubstring(self, s: str) -> str:
i = 0
j = 1
k = 0 # the number of the same letters of s[i..n) and s[j..n)
while j + k < len(s):
if s[i + k] == s[j + k]:
k += 1
elif s[i + k] > s[j + k]:
# Skip s[j..j + k) and advance to s[j + k + 1] to find a possible
# lexicographically larger substring since s[i..i + k) == s[j..j + k)
# and s[i + k] > s[j + k).
j = j + k + 1
k = 0
else:
# Skip s[i..i + k) and advance to s[i + k + 1] or s[j] to find a
# possible lexicographically larger substring since
# s[i..i + k) == s[j..j + k) and s[i + k] < s[j + k).
# Note that it's unnecessary to explore s[i + k + 1..j) if
# i + k + 1 < j since they are already explored by j.
i = max(i + k + 1, j)
j = i + 1
k = 0
return s[i:]