What is the time complexity of Remove One Element to Make the Array Strictly Increasing?

The optimal solution for Remove One Element to Make the Array Strictly Increasing runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only traverse the array once, and the space complexity is O(1) since we use a fixed amount of extra space.

What is the brute force solution for Remove One Element to Make the Array Strictly Increasing?

The brute force approach for Remove One Element to Make the Array Strictly Increasing is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible array formed by removing one element from the original array. If any of these arrays is strictly increasing, we return true. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Remove One Element to Make the Array Strictly Increasing?

Remove One Element to Make the Array Strictly Increasing uses the following concepts: Array. The recommended patterns to study are: Two Pointers, Array.

What are the interview tips for Remove One Element to Make the Array Strictly Increasing?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases and how they might affect your solution. Explain your thought process clearly while coding; it shows your understanding and helps the interviewer follow along.

What are common mistakes when solving Remove One Element to Make the Array Strictly Increasing?

Students often forget to check the edge cases where removing an element doesn't resolve the violation. Some may not account for the scenario where the violation occurs at the start or end of the array.

#1909

Remove One Element to Make the Array Strictly Increasing

Easy

ArrayTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of checking all possible arrays, we can traverse the array once and count the number of violations of the strictly increasing condition. If there’s more than one violation, we can't fix it by removing just one element.

⚙️

Algorithm

3 steps

1Step 1: Initialize a counter for violations of the strictly increasing condition.
2Step 2: Traverse the array and check for violations (where nums[i] <= nums[i - 1]).
3Step 3: If there is more than one violation, return false. If there is one or none, return true.

solution.py10 lines

1def canBeMadeIncreasing(nums):
2    count = 0
3    for i in range(1, len(nums)):
4        if nums[i] <= nums[i - 1]:
5            count += 1
6            if count > 1:
7                return False
8            if i > 1 and nums[i] <= nums[i - 2] and i + 1 < len(nums) and nums[i + 1] <= nums[i - 1]:
9                return False
10    return True

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once, and the space complexity is O(1) since we use a fixed amount of extra space.

1Removing one element can fix at most one violation of the strictly increasing condition.
2If there are multiple violations, the array cannot be made strictly increasing by removing just one element.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.