What is the time complexity of Longest Subarray of 1's After Deleting One Element?

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

What data structures are used to solve Longest Subarray of 1's After Deleting One Element?

This problem uses Array, Dynamic Programming, Sliding Window. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Longest Subarray of 1's After Deleting One Element?

Insight 1: The core idea is to use a sliding window approach to efficiently explore all possible subarrays after deleting one element. Insight 2: Maintain a count of zeros within the current window. The window can expand until it contains more than one zero, then the left boundary of the window needs to move until it contains at most one zero.

Which companies ask LeetCode 1493. Longest Subarray of 1's After Deleting One Element in interviews?

This problem is frequently asked at Tinkoff, VK, Yandex.

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Longest Subarray of 1's After Deleting One Element - LeetCode 1493 Solution

Longest Subarray of 1's After Deleting One Element - Complete Solution Guide

Longest Subarray of 1's After Deleting One Element is LeetCode problem 1493, a Medium 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 binary array nums , you should delete one element from it. Return the size of the longest non-empty subarray containing only 1 's in the resulting array . Return 0 if there is no such subarray. Example 1: Input: nums = [1,1,0,1] Output: 3 Explanation: After deleting the number in position 2, [1,1,1] contains 3 numbers with value of 1's. Example 2: Input: nums = [0,1,1,1,0,1,1,0,1] Output: 5 Explanation: After deleting the number in position 4, [0,1,1,1,1,1,0,1] longest subarray with valu

Detailed Explanation

The problem asks us to find the length of the longest subarray containing only 1's in a given binary array (array of 0s and 1s), after deleting *exactly one* element. The element to be deleted can be a 0 or a 1. If deleting an element does not produce any valid non-empty subarray of 1s, we return 0. The constraints specify the array length is between 1 and 10^5, and each element is either 0 or 1. The goal is to maximize the length of the subarray of 1s after the deletion.

Solution Approach

The solution uses a sliding window approach. We maintain a window with at most one zero. We iterate through the array, expanding the window to the right. If the number of zeros in the window exceeds one, we shrink the window from the left until the number of zeros is at most one again. We keep track of the maximum length of the window at each step. Finally, we return the maximum length found minus 1 since the problem states that one element must be removed.

Step-by-Step Algorithm

Step 1: Initialize `left = 0`, `zeros_count = 0`, and `max_length = 0`.
Step 2: Iterate through the array from `right = 0` to `n - 1`. This is the expansion part of sliding window.
Step 3: If `nums[right]` is 0, increment `zeros_count`.
Step 4: While `zeros_count > 1`, shrink the window from the left. If `nums[left]` is 0, decrement `zeros_count`. Increment `left`.
Step 5: Update `max_length` with the maximum of `max_length` and `right - left + 1`. This is the length of the current window.
Step 6: After the loop finishes, return `max_length - 1`. This is because we need to remove one element to meet the criteria.

Key Insights

Insight 1: The core idea is to use a sliding window approach to efficiently explore all possible subarrays after deleting one element.
Insight 2: Maintain a count of zeros within the current window. The window can expand until it contains more than one zero, then the left boundary of the window needs to move until it contains at most one zero.
Insight 3: After iterating through the array, the resulting length is the 'max_length' minus 1 because we have counted an unnecessary element.
Insight 4: Need to take care when all elements in the array are 1s. When all elements are 1s, then the longest subarray is `n-1`.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Sliding Window.

Companies

Asked at: Tinkoff, VK, Yandex.

Longest Subarray of 1's After Deleting One Element | LeetCode 1493

Medium

Array Dynamic Programming Sliding Window

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int longestSubarray(int[] nums) {
        int n = nums.length;
        int left = 0;
        int zerosCount = 0;
        int maxLength = 0;

        for (int right = 0; right < n; right++) {
            if (nums[right] == 0) {
                zerosCount++;
            }

            while (zerosCount > 1) {
                if (nums[left] == 0) {
                    zerosCount--;
                }
                left++;
            }

            maxLength = Math.max(maxLength, right - left + 1);
        }

        return maxLength - 1;
    }
}

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Problem Info

Description

Problem 1493: Longest Subarray of 1's After Deleting One Element - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Sliding Window

Companies

TinkoffVKYandex

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:1493

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

6 steps

1
Step 1: Initialize `left = 0`, `zeros_count = 0`, and `max_length = 0`.
2
Step 2: Iterate through the array from `right = 0` to `n - 1`. This is the expansion part of sliding window.
3
Step 3: If `nums[right]` is 0, increment `zeros_count`.
4
Step 4: While `zeros_count > 1`, shrink the window from the left. If `nums[left]` is 0, decrement `zeros_count`. Increment `left`.
5
Step 5: Update `max_length` with the maximum of `max_length` and `right - left + 1`. This is the length of the current window.
6
Step 6: After the loop finishes, return `max_length - 1`. This is because we need to remove one element to meet the criteria.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because we iterate through the array once with the `right` pointer, and the `left` pointer can only move from 0 to n, so the inner while loop also runs at most n times in total. Therefore, the overall time complexity is O(n). The space complexity is O(1) because we only use a few constant extra variables (`left`, `zeros_count`, `max_length`).

📖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

Prefix and Suffix Arrays

Calculate prefix and suffix arrays of contiguous 1s. Then, for each index `i`, calculate the sum of the suffix of `i+1` and prefix of `i-1`. Iterate through all possible deleted indices and find the maximum sum.

Time: O(n)Space: O(n)

Trade-offs: This approach uses more space (O(n)) but can be more readable and easier to understand for some. The sliding window approach is more space-efficient.

✨

Key Insights

1Insight 1: The core idea is to use a sliding window approach to efficiently explore all possible subarrays after deleting one element.
2Insight 2: Maintain a count of zeros within the current window. The window can expand until it contains more than one zero, then the left boundary of the window needs to move until it contains at most one zero.
3Insight 3: After iterating through the array, the resulting length is the 'max_length' minus 1 because we have counted an unnecessary element.
4Insight 4: Need to take care when all elements in the array are 1s. When all elements are 1s, then the longest subarray is `n-1`.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: Array with all 1s, e.g., `[1, 1, 1]`. The code correctly returns `n - 1` = 2. The constraint states we must remove one element, so the result will always be less than the array length in this case.
Edge case 2: Array with all 0s, e.g., `[0, 0, 0]`. The code handles this correctly because `max_length` remains 0 throughout, resulting in a return value of `0 - 1 = -1`. Since the problem statement specifies that we return 0 if there is no such subarray, this is incorrect. However, the Python and Java codes actually return `-1 - 1 = -2`. It should be corrected to handle this edge case where `max_length` is zero by checking if the array only contains zeros before returning, and setting the return to 0 instead. (Or, a `max` function could be used to ensure it's never less than zero). The C code handles this fine because the length comparison is performed only when `max_length` is smaller than `(right - left + 1)`. Therefore the answer returned in C code will be `-1`, but in python or java this would return a value smaller than -1.
Edge case 3: Array with a single element. The code returns 0, which is correct because one element has to be deleted.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the array contains only 1s. The problem requires deleting one element, even if all elements are 1s.

✗Common mistake 2: Not properly shrinking the window from the left when the number of zeros exceeds one. This can lead to incorrect window sizes and an inaccurate result.

✗Common mistake 3: Forgetting to subtract 1 from the final result, as one element *must* be removed from the original array

✗Common mistake 4: Assuming that if the `max_length` is zero, the answer must be 0. Instead you must check to verify that every value in array is equal to zero

✅

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: Analyzing network traffic data to find the longest period of uninterrupted service after a single server failure (represented by a 0).

🏦Real-world scenario 2: Finding the longest contiguous sequence of successful operations in a manufacturing process after allowing for one error or malfunction (represented by a 0).

🎮Real-world scenario 3: Determining the longest streak of passing grades after dropping the lowest test score.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: July 24, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

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

@TechLeadJul 6

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

@DevExpert92Jun 29

❓

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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Longest Subarray of 1's After Deleting One Element"?

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

🤖 Need help with this problem?