Introduction

In this blog post, we will explore how to solve the LeetCode problem "1909" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. The array nums is strictly increasing if nums[i - 1] < nums[i] for each index (1 <= i < nums.length). Example 1: Input: nums = [1,2,10,5,7] Output: true Explanation: By removing 10 at index 2 from nums, it becomes [1,2,5,7]. [1,2,5,7] is strictly increasing, so return true. Example 2: Input: nums = [2,3,1,2] Output: false Explanation: [3,1,2] is the result of removing the element at index 0. [2,1,2] is the result of removing the element at index 1. [2,3,2] is the result of removing the element at index 2. [2,3,1] is the result of removing the element at index 3. No resulting array is strictly increasing, so return false. Example 3: Input: nums = [1,1,1] Output: false Explanation: The result of removing any element is [1,1]. [1,1] is not strictly increasing, so return false. Constraints: 2 <= nums.length <= 1000 1 <= nums[i] <= 1000

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• Approach:

  • Iterate through the array and for each element, temporarily "remove" it by skipping it during a strict increasing check.
  • Implement a helper function to determine if an array is strictly increasing.
  • Return True if removing any one element results in a strictly increasing array, otherwise False.

• Complexity:

  • Runtime: O(n^2), where n is the length of nums. Storage: O(1)

Code

    def canBeIncreasing(nums):
    """
    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.

    :param nums: List[int]
    :return: bool
    """

    def is_strictly_increasing(arr):
        """Helper function to check if an array is strictly increasing."""
        for i in range(1, len(arr)):
            if arr[i - 1] >= arr[i]:
                return False
        return True

    n = len(nums)

    # Check if original array is already strictly increasing
    if is_strictly_increasing(nums):
        return True

    for i in range(n):
        temp_nums = []
        for j in range(n):
            if i != j:
                temp_nums.append(nums[j])

        if is_strictly_increasing(temp_nums):
            return True

    return False