BinarySearch 0093 A Strictly Increasing Linked List

Problem statement

https://binarysearch.com/problems/A-Strictly-Increasing-Linked-List/

Solution

Just traverse our linked list and if we have next element less or equal than current, return False. Return True in the end if we reached it.

Complexity

It is O(n) for time and O(1) for space.

Code

class Solution:
    def solve(self, head):
        while head.next:
            if head.next.val <= head.val:
                return False
            head = head.next
        return True