What is the time complexity of Max Consecutive Ones III?

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

What data structures are used to solve Max Consecutive Ones III?

This problem uses Array, Binary Search, Sliding Window, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Max Consecutive Ones III?

Insight 1: The core idea is to use a sliding window approach to maintain a subarray where the number of 0s is always less than or equal to `k`. Insight 2: We need to efficiently track the number of 0s within the sliding window as it expands and contracts.

Which companies ask LeetCode 1004. Max Consecutive Ones III in interviews?

This problem is frequently asked at Expedia, Roku, SAP, Sigmoid, Snap and 3 more companies.

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Max Consecutive Ones III - LeetCode 1004 Solution

Max Consecutive Ones III - Complete Solution Guide

Max Consecutive Ones III is LeetCode problem 1004, 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 and an integer k , return the maximum number of consecutive 1 's in the array if you can flip at most k 0 's. Example 1: Input: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2 Output: 6 Explanation: [1,1,1,0,0, 1 ,1,1,1,1, 1 ] Bolded numbers were flipped from 0 to 1. The longest subarray is underlined. Example 2: Input: nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3 Output: 10 Explanation: [0,0, 1,1, 1 , 1 ,1,1,1, 1 ,1,1 ,0,0,0,1,1,1,1] Bolded numbers were flipped from 0 t

Detailed Explanation

The problem 'Max Consecutive Ones III' asks us to find the length of the longest contiguous subarray within a given binary array (`nums`) that contains only 1s, allowing for at most `k` flips of 0s to 1s. In essence, we want to maximize the length of a window where the number of 0s is no more than `k`. The input is the binary array `nums` and an integer `k`. The output is a single integer representing the maximum length of such a subarray. The constraint is that we can flip at most `k` zeros to ones.

Solution Approach

The solution uses a sliding window approach. We maintain two pointers, `left` and `right`, representing the start and end of the window. We expand the window by incrementing `right`. If we encounter a 0, we increment a `zeros_count`. If `zeros_count` exceeds `k`, we shrink the window by incrementing `left` until `zeros_count` is no longer greater than `k`. While shrinking the window, if the `left` pointer lands on a 0, we decrement `zeros_count`. The maximum length of the window encountered during this process is the answer.

Step-by-Step Algorithm

Step 1: Initialize `left` to 0, `max_len` to 0, and `zeros_count` to 0.
Step 2: Iterate through the `nums` array with the `right` pointer from 0 to `len(nums) - 1`.
Step 3: If `nums[right]` is 0, increment `zeros_count`.
Step 4: While `zeros_count` is greater than `k`, check if `nums[left]` is 0. If so, decrement `zeros_count`. Increment `left` to shrink the window.
Step 5: Update `max_len` to be the maximum of `max_len` and the current window length (`right - left + 1`).
Step 6: After the loop finishes, return `max_len`.

Key Insights

Insight 1: The core idea is to use a sliding window approach to maintain a subarray where the number of 0s is always less than or equal to `k`.
Insight 2: We need to efficiently track the number of 0s within the sliding window as it expands and contracts.
Insight 3: The problem can be solved in O(n) time complexity with O(1) space complexity because we can adjust the window as needed without needing to store all possible subarrays or extra data structures related to the input array's size.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Sliding Window, Prefix Sum.

Companies

Asked at: Expedia, Roku, SAP, Sigmoid, Snap, Walmart Labs, Yandex, tcs.

Max Consecutive Ones III | LeetCode 1004

Medium

Array Binary Search Sliding Window Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int longestOnes(int[] nums, int k) {
        int left = 0;
        int max_len = 0;
        int zeros_count = 0;

        for (int right = 0; right < nums.length; right++) {
            if (nums[right] == 0) {
                zeros_count++;
            }

            while (zeros_count > k) {
                if (nums[left] == 0) {
                    zeros_count--;
                }
                left++;
            }
            
            max_len = Math.max(max_len, right - left + 1);
            
        }
        return max_len;
    }
}

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Problem Info

Description

Problem 1004: Max Consecutive Ones III - Detailed explanation and solution approaches.

Topics

Array Binary Search Sliding Window Prefix Sum

Companies

ExpediaRokuSAPSigmoidSnapWalmart LabsYandextcs

Stats

Difficulty:Medium

Languages:4

Companies:8

Problem #:1004

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` to 0, `max_len` to 0, and `zeros_count` to 0.
2
Step 2: Iterate through the `nums` array with the `right` pointer from 0 to `len(nums) - 1`.
3
Step 3: If `nums[right]` is 0, increment `zeros_count`.
4
Step 4: While `zeros_count` is greater than `k`, check if `nums[left]` is 0. If so, decrement `zeros_count`. Increment `left` to shrink the window.
5
Step 5: Update `max_len` to be the maximum of `max_len` and the current window length (`right - left + 1`).
6
Step 6: After the loop finishes, return `max_len`.

💡

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 both `left` and `right` pointers traverse the array at most once. The inner `while` loop might seem like it increases the complexity, but because `left` never moves backward, the total number of times the `while` loop executes is bounded by n. The space complexity is O(1) because the algorithm uses only a fixed number of variables, regardless of the input array's size.

📖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 Sum and Binary Search

We can calculate a prefix sum array that stores the cumulative number of zeros. Then, for each possible subarray length, we can use binary search to find the leftmost index where flipping the zeros in that subarray would exceed k. This approach is more complex.

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

Trade-offs: The sliding window approach is more efficient (O(n)) and simpler to implement than prefix sum with binary search (O(n log n)). The prefix sum approach is useful when similar queries need to be performed repeatedly on the same array.

✨

Key Insights

1Insight 1: The core idea is to use a sliding window approach to maintain a subarray where the number of 0s is always less than or equal to `k`.
2Insight 2: We need to efficiently track the number of 0s within the sliding window as it expands and contracts.
3Insight 3: The problem can be solved in O(n) time complexity with O(1) space complexity because we can adjust the window as needed without needing to store all possible subarrays or extra data structures related to the input array's size.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sliding Window

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

Edge Cases

Edge case 1: `k` is 0. In this case, the algorithm correctly finds the longest contiguous subarray of 1s without flipping any 0s.
Edge case 2: `k` is greater than or equal to the number of 0s in `nums`. In this case, the algorithm correctly returns the length of the entire `nums` array.
Edge case 3: `nums` is an empty array. Although the problem statement implies a non-empty array, a robust solution should handle an empty array input. The given solution will work since the loop won't execute and 0 will be returned.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the condition `zeros_count > k`. It's important to shrink the window from the `left` until the condition is no longer true.

✗Common mistake 2: Not updating `max_len` after each iteration of the outer loop. This can lead to an incorrect result, especially if the longest subarray is at the beginning of the array.

✗Common mistake 3: Forgetting to decrement `zeros_count` when shrinking the window and encountering a zero at the `left` pointer.

✅

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: In data transmission, this algorithm can be used to find the longest consecutive sequence of successfully transmitted packets, allowing for a certain number of corrupted packets (`k`) that can be recovered or retransmitted.

🏦Real-world scenario 2: In DNA sequencing, this algorithm can be used to find the longest contiguous region of high-quality reads, allowing for a certain number of low-quality reads (`k`) that can be corrected or ignored.

💡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: September 13, 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 13

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

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Max Consecutive Ones III"?

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 binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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