How do you solve Max Consecutive Ones III on LeetCode?

Max Consecutive Ones III (LeetCode #1004) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using a sliding window allows us to efficiently manage the count of zeros without needing nested loops.

What is the brute force solution for Max Consecutive Ones III?

The brute force approach for Max Consecutive Ones III is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible subarray to find the maximum length of consecutive 1's that can be formed by flipping at most k 0's. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Max Consecutive Ones III?

Max Consecutive Ones III uses the following concepts: Array, Binary Search, Sliding Window, Prefix Sum. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Max Consecutive Ones III?

Always think about how you can reduce the problem size with techniques like sliding windows or two pointers. Practice explaining your thought process as you code; it helps clarify your understanding. Be mindful of edge cases and test your solution against them.

What are common mistakes when solving Max Consecutive Ones III?

Not properly managing the left pointer when the count of zeros exceeds k. Failing to consider edge cases, such as when k is 0 or when the array is all 1's.

#1004

Max Consecutive Ones III

Medium

ArrayBinary SearchSliding WindowPrefix SumSliding WindowTwo Pointers

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)

The optimal approach uses the sliding window technique to maintain a window of valid elements (1's and up to k flipped 0's). This allows us to efficiently find the maximum length of consecutive 1's without checking every subarray.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers (left and right) and a variable to count zeros in the current window.
2Step 2: Expand the right pointer to include more elements in the window.
3Step 3: If the count of zeros exceeds k, move the left pointer to reduce the window size until the count of zeros is k or less.
4Step 4: Update the maximum length of the window whenever the condition is satisfied.

solution.py15 lines

1# Full working Python code
2
3def longestOnes(nums, k):
4    left = 0
5    zero_count = 0
6    max_length = 0
7    for right in range(len(nums)):
8        if nums[right] == 0:
9            zero_count += 1
10        while zero_count > k:
11            if nums[left] == 0:
12                zero_count -= 1
13            left += 1
14        max_length = max(max_length, right - left + 1)
15    return max_length

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once with two pointers. The space complexity is O(1) since we are using a constant amount of extra space.

1Using a sliding window allows us to efficiently manage the count of zeros without needing nested loops.
2The maximum length is updated only when the current window is valid, which avoids unnecessary calculations.

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