Leetcode simple problem delete an element to strictly increment the array

subject

I'll give you a subscript from 0 The starting array of integers nums , If just Delete One After the element , Array Strictly increasing , So please go back to true , Otherwise return to false . If the array itself is strictly incremented , Please return, too true .
Array nums yes Strictly increasing For the definition of ： For any subscript 1 <= i < nums.length All satisfied with nums[i - 1] < nums[i] .
Example 1：
Input ：nums = [1,2,10,5,7]
Output ：true
explain ： from nums Remove the subscript from the 2 Situated 10 , obtain [1,2,5,7] .
[1,2,5,7] It's strictly incremental , So back true .
Example 2：
Input ：nums = [2,3,1,2]
Output ：false
explain ：
[3,1,2] Is to delete the subscript 0 The result of the element .
[2,1,2] Is to delete the subscript 1 The result of the element .
[2,3,2] Is to delete the subscript 2 The result of the element .
[2,3,1] Is to delete the subscript 3 The result of the element .
No result array is strictly incremented , So back false .
Example 3：
Input ：nums = [1,1,1]
Output ：false
explain ： The result of deleting any element is [1,1] .
[1,1] Not strictly incremental , So back false .
Example 4：
Input ：nums = [1,2,3]
Output ：true
explain ：[1,2,3] It has been strictly increasing , So back true .
Tips ：
2 <= nums.length <= 1000
1 <= nums[i] <= 1000
source ： Power button （LeetCode）

Their thinking

   A simple direction is to traverse the array first , Find the point where the elements in the array do not conform to strict increment , Then modify the array by analyzing the situation , Of course, you can only delete it once , Finally, another round of judgment .

class Solution:
    def canBeIncreasing(self, nums: List[int]) -> bool:
        i=1
        n=len(nums)
        while i<len(nums):
            if nums[i]>nums[i-1]:
                i+=1
            else:  # If the current point makes the array not strictly single increment 
                if i-2>=0:  # Check the second point before the current element 
                    if nums[i]>nums[i-2]: # If the current point and the previous second point form a strict increment , Then delete the first point in front of the current element 
                        del nums[i-1]
                    else:  # Otherwise, delete the current element 
                        del nums[i]
                else:  # If there is only one element in front , Then delete the current point 
                    del nums[i-1]
                break
        if n==len(nums):  # There was no deletion , Strict single increment of the original array 
            return True
        for i in range(1,len(nums)): # Whether to strictly add after a deletion 
            if nums[i]<=nums[i-1]:
                return False
        return True