Leetcode #1909: Remove One Element to Make the Array Strictly Increasing
In this guide, we solve Leetcode #1909 Remove One Element to Make the Array Strictly Increasing in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Array
Intuition
The constraints allow a direct scan that keeps only the essential state.
By translating the requirements into a clean loop, the logic stays easy to reason about.
Approach
Iterate through the data once, updating the state needed to compute the answer.
Return the final state after the traversal is complete.
Steps:
- Parse the input.
- Iterate and update state.
- Return the computed answer.
Example
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.
Python Solution
class Solution:
def canBeIncreasing(self, nums: List[int]) -> bool:
def check(k: int) -> bool:
pre = -inf
for i, x in enumerate(nums):
if i == k:
continue
if pre >= x:
return False
pre = x
return True
i = 0
while i + 1 < len(nums) and nums[i] < nums[i + 1]:
i += 1
return check(i) or check(i + 1)
Complexity
The time complexity is , where is the length of the array . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.