Check If Reversing a Sub-Array Makes the Array Sorted in Python

Suppose we have an array called nums with unique elements. We have to check whether the array will be sorted or not after reversing one sub-array of it. If the array is already sorted, then also return true.

So, if the input is like nums = [4,6,27,25,15,9,37,42], then the output will be True because if we reverse [9,15,25,27], then the array will be sorted.

To solve this, we will follow these steps −

• n := size of nums
• if array has only one element then return True
• i := 1
• for i in range 1 to n - 1, do
  • if nums[i - 1] < nums[i], then
    • if i is same as n, then return true, otherwise come out from loop
• j := i
• while j < n and nums[j] < nums[j - 1], do
  • if i > 1 and nums[j] < nums[i - 2], then return false
  • j := j + 1
• if j is same as n, then return True
• k := j
• if nums[k] < nums[i - 1], then return False
• while k > 1 and k < n, do
  • if nums[k] < nums[k - 1], then return False
  • k := k + 1
• return True

Let us see the following implementation to get better understanding −

Example Code

Live Demo

def solve(nums):
   n = len(nums)
   if n == 1:
      return True
 
   i = 1
   for i in range(1, n):
      if nums[i - 1] < nums[i] :
         if i == n:
           return True
      else:
         break
   j = i
   
   while j < n and nums[j] < nums[j - 1]:
      if i > 1 and nums[j] < nums[i - 2]:
         return False
      j += 1
 
   if j == n:
      return True
 
   k = j
   if nums[k] < nums[i - 1]:
      return False
 
   while k > 1 and k < n:
      if nums[k] < nums[k - 1]:
         return False
      k += 1
   return True

nums = [4,6,27,25,15,9,37,42]
print(solve(nums))

Input

[4,6,27,25,15,9,37,42]

Output

True