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 true if and only if the given array A is monotonic.
Example 1:
Input: [1,2,2,3] Output: true
Example 2:
Input: [6,5,4,4] Output: true
Example 3:
Input: [1,3,2] Output: false
Example 4:
Input: [1,2,4,5] Output: true
Example 5:
Input: [1,1,1] Output: true
Note:
1 <= A.length <= 50000-100000 <= A[i] <= 100000
The description was taken from https://leetcode.com/problems/monotonic-array/.
#O(N) Time, O(1) Spaceclass Solution: def isMonotonic(self, A: List[int]) -> bool: increasing = decreasing = True for i in range(1, len(A)): if A[i] > A[i-1]: decreasing = False elif A[i] < A[i-1]: increasing = False return increasing or decreasingWith a problem like this, where there are only a couple of possible use cases like whether the array is increasing or decreasing, many times we can just make a conditional statement for each use case and approach the question that way.
We'll have one conditional statement to check if the array is decreasing, one to see if it's increasing, and continue to update these values.
In the end, we will see one of the conditions remained true.
In the description, it specifies the array is monotonic if it is strictly increasing or decreasing so we will create these two variables to store the answers.
increasing = decreasing = True
We'll then traverse the array starting from the second index and see if the element behind us is higher or lower.
for i in range(1, len(A)):
If it is lower, we know we aren't decreasing.
if A[i] > A[i-1]: decreasing = False
If it is higher, we know we aren't increasing.
elif A[i] < A[i-1]: increasing = False
We'll return whether or not one of these two conditions remained true once we reached the end of the array.
return increasing or decreasing