Python3 Program for Shortest Un-ordered Subarray
An array is given of n length, and problem is that we have to find the length of shortest unordered {neither increasing nor decreasing} sub array in given array.
Examples:
Input : n = 5
7 9 10 8 11
Output : 3
Explanation : 9 10 8 unordered sub array.
Input : n = 5
1 2 3 4 5
Output : 0
Explanation : Array is in increasing order.
The idea is based on the fact that size of shortest subarray would be either 0 or 3. We have to check array element is either increasing or decreasing, if all array elements are in increasing or decreasing, then length of shortest sub array is 0, And if either the array element is not follow the increasing or decreasing then it shortest length is 3.
Python3
def increasing(a, n):
for i in range ( 0 , n - 1 ):
if (a[i] > = a[i + 1 ]):
return False
return True
def decreasing(a, n):
for i in range ( 0 , n - 1 ):
if (a[i] < a[i + 1 ]):
return False
return True
def shortestUnsorted(a, n):
if (increasing(a, n) = = True or
decreasing(a, n) = = True ):
return 0
else :
return 3
ar = [ 7 , 9 , 10 , 8 , 11 ]
n = len (ar)
print (shortestUnsorted(ar, n))
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Time complexity: O(n) where n is the length of the array.
Auxiliary Space: O(1)
Please refer complete article on Shortest Un-ordered Subarray for more details!
Last Updated :
28 May, 2022
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