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Python Program to check if given array is Monotonic

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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




# Check if given array is Monotonic
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
 
 
# Driver program
A = [6, 5, 4, 4]
 
# Print required result
print(isMonotonic(A))


Output

True

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




# Python Program to check if given array is Monotonic
 
# Check if given array is Monotonic
 
 
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)))
 
 
# Driver program
A = [6, 5, 4, 4]
 
# Print required result
print(isMonotonic(A))
 
# This code is written by
# Sanjit_Prasad


Output

True

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
 
# Example usage
arr1 = [1, 2, 3, 4, 5] # True
arr2 = [5, 4, 3, 2, 1] # True
arr3 = [1, 2, 2, 3, 4] # True
arr4 = [1, 2, 3, 4, 5, 4] # False
 
print(isMonotonic(arr1)) # should return True
print(isMonotonic(arr2)) # should return True
print(isMonotonic(arr3)) # should return True
print(isMonotonic(arr4)) # should return False


Output

True
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.

Approach : By using the set to Identify Unique Elements

In this we will check, if the array is monotonic by converting it to a set to identify unique elements and then determining monotonicity by comparing the array against both its ascending and descending sorted versions. If either comparison holds true, the array is considered monotonic.

Below is implementation for the above approach:

Python3




def is_monotonic(arr):
    unique_elements = set(arr)
    increasing = sorted(arr) == arr or sorted(arr, reverse=True) == arr
    return increasing
 
# Driver Code
arr1 = [6, 5, 4, 4]
arr2 = [5, 15, 20, 10]
arr3 = [2, 2, 2, 3]
 
print(is_monotonic(arr1)) 
print(is_monotonic(arr2)) 
print(is_monotonic(arr3)) 


Output

True
False
True

Time Complexity: O(n)
Auxiliary Space: O(n)



Last Updated : 04 Dec, 2023
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