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Suppose we have a lowercase string s. We have to find the minimum numbers of contiguous substrings in which s is divided into parts such that each substring is either non-increasing or non-decreasing. So for example, if the string is like "pqqqr" is a non-decreasing string, and "qqqp" is a non-increasing string.

So, if the input is like s = "pqrsrqp", then the output will be 2, because we can break s like "pqrs" and "rqp".

To solve this, we will follow these steps −

if s is empty, then

return 0

last := s[0]

direction := 1

count := 1

for each char in s, do

if char > last, then

if direction is same as 1, then

direction := 0

otherwise when direction is same as 2, then

direction := 1

count := count + 1

otherwise when char < last, then

if direction is same as 1, then

direction := 2

otherwise when direction is same as 0, then

direction := 1

count := count + 1

last := char

return count

Let us see the following implementation to get better understanding

def solve(s): if not s: return 0 last = s[0] direction = 1 count = 1 for char in s: if char > last: if direction == 1: direction = 0 elif direction == 2: direction = 1 count += 1 elif char < last: if direction == 1: direction = 2 elif direction == 0: direction = 1 count += 1 last = char return count s = "pqrsrqp" print(solve(s))

"pqrsrqp"

2

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