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

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Remove One Element to Make the Array Strictly Increasing?

This problem uses Array. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Remove One Element to Make the Array Strictly Increasing?

Insight 1: The problem can be solved by iterating through the array and removing each element one at a time. Then, check if the resulting array is strictly increasing. Insight 2: A more efficient approach involves a single pass through the array to count the number of violations of the strictly increasing condition. If there's only one violation, or none at all, the array can be made strictly increasing by removing at most one element.

Which companies ask LeetCode 1909. Remove One Element to Make the Array Strictly Increasing in interviews?

This problem is frequently asked at eBay.

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Remove One Element to Make the Array Strictly Increasing - LeetCode 1909 Solution

Remove One Element to Make the Array Strictly Increasing - Complete Solution Guide

Remove One Element to Make the Array Strictly Increasing is LeetCode problem 1909, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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: n

Detailed Explanation

The problem asks whether a given array of integers can be made strictly increasing by removing at most one element. A strictly increasing array means that each element is strictly greater than the previous element (nums[i] > nums[i-1] for all i > 0). The input is a 0-indexed integer array `nums`, and the output is a boolean value: `true` if the array can be made strictly increasing by removing one element, or `false` otherwise. The array's length is between 2 and 1000, and elements are between 1 and 1000.

Solution Approach

The provided Python, Java, and C++ solutions employ different strategies. The Python and Java solutions directly implement the brute-force approach: they iterate through the array, remove each element, and check if the remaining array is strictly increasing. The C++ and C solutions are more optimized, performing a single pass to check for violations and cleverly handling edge cases, without creating new arrays in each iteration.

Step-by-Step Algorithm

Step 1 (Brute-force): Iterate through each element of the input array `nums`.
Step 2 (Brute-force): Create a temporary array `temp` excluding the current element.
Step 3 (Brute-force): Check if `temp` is strictly increasing. If it is, return `true`.
Step 4 (Brute-force): If no temporary array is strictly increasing after iterating through all elements, return `false`.
Step 1 (Optimized): Iterate through the array, keeping track of the number of violations of the strictly increasing condition.
Step 2 (Optimized): If a violation is found, check if removing the current element or the previous element would resolve the issue. This involves a conditional check involving the element before and after the violation.
Step 3 (Optimized): If the number of violations exceeds 1, return `false`; otherwise, return `true`.

Key Insights

Insight 1: The problem can be solved by iterating through the array and removing each element one at a time. Then, check if the resulting array is strictly increasing.
Insight 2: A more efficient approach involves a single pass through the array to count the number of violations of the strictly increasing condition. If there's only one violation, or none at all, the array can be made strictly increasing by removing at most one element.
Insight 3: Edge cases need consideration. For example, an array with all identical elements cannot be made strictly increasing, even after removing one element.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array.

Companies

Asked at: eBay.

Remove One Element to Make the Array Strictly Increasing | LeetCode 1909

Easy

Array

Time: O(n^2)Space: O(n)

Solution Code

Available in:

class Solution {
    public boolean canBeIncreasing(int[] nums) {
        for (int i = 0; i < nums.length; i++) {
            int[] temp = new int[nums.length - 1];
            int k = 0;
            for (int j = 0; j < nums.length; j++) {
                if (j != i) {
                    temp[k++] = nums[j];
                }
            }
            boolean increasing = true;
            for (int j = 1; j < temp.length; j++) {
                if (temp[j] <= temp[j - 1]) {
                    increasing = false;
                    break;
                }
            }
            if (increasing) {
                return true;
            }
        }
        return false;
    }
}

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Problem Info

Description

Problem 1909: Remove One Element to Make the Array Strictly Increasing - Detailed explanation and solution approaches.

Topics

Array

Companies

eBay

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1909

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1 (Brute-force): Iterate through each element of the input array `nums`.
2
Step 2 (Brute-force): Create a temporary array `temp` excluding the current element.
3
Step 3 (Brute-force): Check if `temp` is strictly increasing. If it is, return `true`.
4
Step 4 (Brute-force): If no temporary array is strictly increasing after iterating through all elements, return `false`.
5
Step 1 (Optimized): Iterate through the array, keeping track of the number of violations of the strictly increasing condition.
6
Step 2 (Optimized): If a violation is found, check if removing the current element or the previous element would resolve the issue. This involves a conditional check involving the element before and after the violation.
7
Step 3 (Optimized): If the number of violations exceeds 1, return `false`; otherwise, return `true`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The brute-force approach creates a new array in each iteration of the outer loop, resulting in O(n^2) time and O(n) space. The optimized approach avoids array creation and has only one loop, resulting in O(n) time and O(1) space complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Binary Search (Not directly applicable)

Binary search is not directly applicable here since we're not searching for a specific value but rather checking a property across the entire array.

Time: N/ASpace: N/A

Trade-offs: Not suitable for this problem

Dynamic Programming (Not efficient)

A DP approach could potentially track the possibilities but would likely be less efficient than the single-pass approach.

Time: O(n^2)Space: O(n)

Trade-offs: Less efficient than the optimized approach, adding complexity without significant benefit

✨

Key Insights

1Insight 1: The problem can be solved by iterating through the array and removing each element one at a time. Then, check if the resulting array is strictly increasing.
2Insight 2: A more efficient approach involves a single pass through the array to count the number of violations of the strictly increasing condition. If there's only one violation, or none at all, the array can be made strictly increasing by removing at most one element.
3Insight 3: Edge cases need consideration. For example, an array with all identical elements cannot be made strictly increasing, even after removing one element.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: An array that is already strictly increasing. The solution should return `true`.
Edge case 2: An array with all identical elements. The solution should return `false`.
Edge case 3: An array with only one violation. The solution should check if removing either the violating element or the previous element makes it strictly increasing.
Edge case 4: An array with multiple non-consecutive violations where removing one element may not suffice to create a strictly increasing array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not properly handling the edge cases (e.g., already strictly increasing array, array with all identical elements).

✗Common mistake 2: Failing to consider that removing the element before the violation might fix the problem in the optimized approach.

✗Common mistake 3: Incorrectly implementing the strict increase check in the inner loop, leading to logic errors.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data validation: Ensuring that a sequence of measurements or data points follows a strictly increasing trend (e.g., stock prices, sensor readings). If a small anomaly is detected, removing it might be acceptable for data analysis.

🏦Real-world scenario 2: Signal processing: Identifying and correcting minor glitches or anomalies in a signal sequence. If a single aberrant data point is found, removing it might be a reasonable approach to noise reduction.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 18, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Remove One Element to Make the Array Strictly Increasing"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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