Given an array A containing n integers. The task is to check whether the array is Monotonic or not. An array is monotonic if it is either monotone increasing or monotone decreasing. An array A is monotone increasing if for all i <= j, A[i] <= A[j].
An array A is monotone decreasing if for all i <= j, A[i] >= A[j].
Return Type: Boolean value, “True” if the given array A is monotonic else return “False” (without quotes).
Examples:
Input : 6 5 4 4
Output : true
Input : 5 15 20 10
Output : false
Approach : Using extend() and sort()
- First copy the given array into two different arrays using extend()
- Sort the first array in ascending order using sort()
- Sort the second array in descending order using sort(reverse=True)
- If the given array is equal to any of the two arrays then the array is monotonic
Python3
def isMonotonic(A):
x, y = [], []
x.extend(A)
y.extend(A)
x.sort()
y.sort(reverse = True )
if (x = = A or y = = A):
return True
return False
A = [ 6 , 5 , 4 , 4 ]
print (isMonotonic(A))
|
Time Complexity: O(N*logN), where N is the length of the array.
Auxiliary space: O(N), extra space is required for lists x and y.
Approach:
An array is monotonic if and only if it is monotone increasing, or monotone decreasing. Since p <= q and q <= r implies p <= r. So we only need to check adjacent elements to determine if the array is monotone increasing (or decreasing), respectively. We can check each of these properties in one pass.
To check whether an array A is monotone increasing, we’ll check A[i] <= A[i+1] for all i indexing from 0 to len(A)-2. Similarly we can check for monotone decreasing where A[i] >= A[i+1] for all i indexing from 0 to len(A)-2.
Note: Array with single element can be considered to be both monotonic increasing or decreasing, hence returns “True“.
Below is the implementation of the above approach:
Python3
def isMonotonic(A):
return ( all (A[i] < = A[i + 1 ] for i in range ( len (A) - 1 )) or
all (A[i] > = A[i + 1 ] for i in range ( len (A) - 1 )))
A = [ 6 , 5 , 4 , 4 ]
print (isMonotonic(A))
|
Time Complexity: O(N), where N is the length of the array.
Auxiliary space: O(1) because it is using constant space.
Approach 3 – By checking length of the array
Python3
def isMonotonic(arr):
if len (arr) < = 2 :
return True
direction = arr[ 1 ] - arr[ 0 ]
for i in range ( 2 , len (arr)):
if direction = = 0 :
direction = arr[i] - arr[i - 1 ]
continue
if (direction > 0 and arr[i] < arr[i - 1 ]) or (direction < 0 and arr[i] > arr[i - 1 ]):
return False
return True
arr1 = [ 1 , 2 , 3 , 4 , 5 ]
arr2 = [ 5 , 4 , 3 , 2 , 1 ]
arr3 = [ 1 , 2 , 2 , 3 , 4 ]
arr4 = [ 1 , 2 , 3 , 4 , 5 , 4 ]
print (isMonotonic(arr1))
print (isMonotonic(arr2))
print (isMonotonic(arr3))
print (isMonotonic(arr4))
|
OutputTrue
True
True
False
This program first checks if the length of the array is less than or equal to 2, in which case it returns True. Next, it initializes a variable “direction” to the difference between the first two elements of the array. Then, it iterates through the rest of the array and checks if the current element is greater than or less than the previous element, depending on the direction. If any element does not match the direction, the function returns False. If the function completes the loop and has not returned False, it returns True.
Please note that this program assumes that the input array is a list of integers, if the input array consists of other data types it will not work as expected.
The time complexity of the above program is O(n). This is because the program iterates through the entire array once, and the amount of time it takes to complete the iteration is directly proportional to the number of elements in the array. The program uses a single variable “direction” to store the difference between the first two elements of the array, and a single variable “i” to keep track of the current index during iteration.
The auxiliary space of the above program is O(1). This is because the program only uses a constant amount of extra memory to store the single variable “direction” and the single variable “i”. The program does not create any new data structures or use any recursion, so it does not require any additional memory beyond the input array.