Python | Check for Descending Sorted List
Last Updated :
09 Apr, 2023
The sorted operation of list is essential operation in many application. But it takes best of O(nlogn) time complexity, hence one hopes to avoid this. So, to check if this is required or not, knowing if list is by default reverse sorted or not, one can check if list is sorted or not. Lets discuss various ways this can be achieved.
Method #1 : Naive method The simplest way to check this is run a loop for first element and check if we could find any larger element than it after that element, if yes, the list is not reverse sorted.
Python3
test_list = [ 10 , 8 , 4 , 3 , 1 ]
print ("Original list : " + str (test_list))
flag = 0
i = 1
while i < len (test_list):
if (test_list[i] > test_list[i - 1 ]):
flag = 1
i + = 1
if ( not flag) :
print ("Yes, List is reverse sorted .")
else :
print ("No, List is not reverse sorted .")
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Output :
Original list : [10, 8, 4, 3, 1]
Yes, List is reverse sorted.
Time Complexity: O(n), where n is the length of the list test_list
Auxiliary Space: O(1) additional space is not needed
Method #2 : Using sort() + reverse The new list can be made as a copy of the original list, sorting the new list and comparing with the old list will give us the result if sorting was required to get reverse sorted list or not.
Python3
test_list = [ 10 , 5 , 4 , 3 , 1 ]
print ("Original list : " + str (test_list))
flag = 0
test_list1 = test_list[:]
test_list1.sort(reverse = True )
if (test_list1 = = test_list):
flag = 1
if (flag) :
print ("Yes, List is reverse sorted .")
else :
print ("No, List is not reverse sorted .")
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Output :
Original list : [10, 8, 4, 3, 1]
Yes, List is reverse sorted.
Time Complexity: O(nlogn), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(1), constant extra space is required
Method #3 : Using all() function
The all() function returns True if all elements of the iterable are True. Hence, by comparing each element with its next, we can find if the list is reverse sorted or not.
Python3
test_list = [ 10 , 8 , 4 , 3 , 1 ]
print ( "Original list : " + str (test_list))
if all (test_list[i] > = test_list[i + 1 ] for i in range ( len (test_list) - 1 )):
print ( "Yes, List is reverse sorted." )
else :
print ( "No, List is not reverse sorted." )
|
Output
Original list : [10, 8, 4, 3, 1]
Yes, List is reverse sorted.
Time complexity: O(n)
Auxiliary Space: O(n)
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