Remove One Element To Make The Array Strictly Increasing - Leetcode Solution

Problem statement:

Given a 0-indexed integer array nums, return true if it can be made strictly increasing after removing exactly one element, or false otherwise. If the array is already strictly increasing, return true.

An array nums is strictly increasing if nums[i] < nums[i+1] for all 0 <= i < nums.length - 1. An array of length 1 is trivially strictly increasing.

Solution:

To solve this problem, we need to check if the array is already strictly increasing. If it is, we return true. If not, we need to check if we can remove one element from the array such that the resulting array is strictly increasing.

Let's start by checking if the array is already strictly increasing. To do this, we can iterate through the array and check if each element is less than the next element. If at any point this condition is not met, we move on to the next step.

If the array is not already strictly increasing, we need to check if we can remove one element and make the array strictly increasing. We can do this by iterating through the array and removing one element at a time, and then checking if the resulting array is strictly increasing.

To remove an element from the array, we can use python's del keyword. We can then use a helper function that checks if an array is strictly increasing. If the resulting array is strictly increasing, we return True.

Here is the python code to solve this problem:

def canBeIncreasing(nums: List[int]) -> bool: # check if array is already strictly increasing if is_strictly_increasing(nums): return True

# try to remove one element and see if the resulting array is strictly increasing
for i in range(len(nums)):
    copy = nums[:]
    del copy[i]
    if is_strictly_increasing(copy):
        return True

return False

def is_strictly_increasing(nums: List[int]) -> bool: for i in range(len(nums)-1): if nums[i] >= nums[i+1]: return False return True

In summary, we check if the array is strictly increasing. If it is, we return True. If not, we iterate through the array and remove one element at a time, and then check if the resulting array is strictly increasing. If we find such an array, we return True. If we can't find such an array, we return False. This solution has a time complexity of O(n^2) and a space complexity of O(n).

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