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Max Consecutive Ones III - Leetcode 1004 - Sliding Window

Disclaimer: This is a personal summary and interpretation based on a YouTube video. It is not official material and not endorsed by the original creator. All rights remain with the respective creators.

This document summarizes the key takeaways from the video. I highly recommend watching the full video for visual context and coding demonstrations.

Before You Get Started

  • I summarize key points to help you learn and review quickly.
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Problem Statement

  • Summary: The problem involves a binary array of 0s and 1s, and an integer K. The goal is to find the maximum number of consecutive 1s achievable by flipping at most K 0s to 1s.
  • Key Takeaway/Example: This is a variable-length sliding window problem where you expand and contract a window to maintain validity based on the number of 0s.
  • Link for More Details: Ask AI: Max Consecutive Ones III Problem

Example Walkthrough

  • Summary: Using an example array like [1,1,1,0,0,0,1,1,1,1,0] with K=2, you can flip two 0s to achieve six consecutive 1s, which is better than smaller segments.
  • Key Takeaway/Example: Flipping specific 0s, such as the fourth and fifth positions, creates a longer streak of 1s.
  • Link for More Details: Ask AI: Sliding Window Example for Leetcode 1004

Sliding Window Mechanics

  • Summary: Use two pointers, L and R, to define the window. Expand with R, track the number of 0s, and contract with L if 0s exceed K to keep the window valid. Update the maximum length when valid.
  • Key Takeaway/Example: The window is valid if the number of 0s is <= K. Contract by moving L and decrement 0s count if removing a 0.
  • Link for More Details: Ask AI: Sliding Window Technique in Leetcode 1004

Code Implementation

  • Summary: Initialize max_w = 0, num_zeros = 0, L = 0. Loop through R from 0 to n-1: if nums[R] == 0, increment num_zeros. While num_zeros > K, if nums[L] == 0 decrement num_zeros, then L += 1. Update max_w with max(max_w, R - L + 1).
  • Key Takeaway/Example:
max_w = 0
num_zeros = 0
n = len(nums)
L = 0
for R in range(n):
    if nums[R] == 0:
        num_zeros += 1
    while num_zeros > K:
        if nums[L] == 0:
            num_zeros -= 1
        L += 1
    w = R - L + 1
    max_w = max(max_w, w)
return max_w

Time and Space Complexity

  • Summary: The algorithm runs in O(n) time since both pointers move forward only, and uses O(1) space for variables.
  • Key Takeaway/Example: Pointers L and R traverse the array at most once each, ensuring linear time.
  • Link for More Details: Ask AI: Complexity of Sliding Window in Leetcode 1004

About the summarizer

I'm Ali Sol, a Backend Developer. Learn more: