Problem

Solution

Intuition

Intuitively we know we want to scan the array twice, once to check for an increasing monotonic condition, and once to check for a decreasing monotonic condition.

Approach

Building on the intuition an efficient solution would be able to check for both conditions in one loop as both conditions check for a solution in a left-to-right manner. We can simultaneously look for both conditions since only one or neither condition will be met during the same left-to-right pass.

Complexity

• Time complexity: The time complexity is \(O(n)\) for the loop and \(O(1)\) for all checks within the for loop thus making the time complexity \(O(n)\).

• Space complexity: There are two constant space variables created so the space complexity is \(O(1)\).

Code

class Solution:
    def isMonotonic(self, nums: List[int]) -> bool:
        # Flag to check for increasing monotonic condition
        incFlag = True
        # Flag to check for decreasing monotonic condition
        decFlag = True
        for i in range(1,len(nums)):
            # Fail condition for a decreasing monotonic condition
            if nums[i] > nums[i-1]:
                decFlag = False
            # Fail condition for a decreasing monotonic condition
            if nums[i] < nums[i-1]:
                incFlag = False
        # Only one or neither flag will remain True after traversal
        return decFlag or incFlag