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
# Python3 program to find shortest # subarray which is unsorted# Bool function for checking an array # elements are in increasingdef increasing(a, n): for i in range(0, n - 1): if (a[i] >= a[i + 1]): return False return True# Bool function for checking an array # elements are in decreasingdef decreasing(a, n): for i in range(0, n - 1): if (a[i] < a[i + 1]): return False return Truedef shortestUnsorted(a, n): # increasing and decreasing are two functions. # if function return True value then print # 0 otherwise 3. if (increasing(a, n) == True or decreasing(a, n) == True): return 0 else: return 3# Driver codear = [7, 9, 10, 8, 11] n = len(ar) print(shortestUnsorted(ar, n))# This code is contributed by Smitha Dinesh Semwal. |
Output :
3
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!
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