Introduction

In this blog post, we will explore how to solve the LeetCode problem "995" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a binary array nums and an integer k. A k-bit flip is choosing a subarray of length k from nums and simultaneously changing every 0 in the subarray to 1, and every 1 in the subarray to 0. Return the minimum number of k-bit flips required so that there is no 0 in the array. If it is not possible, return -1. A subarray is a contiguous part of an array. Example 1: Input: nums = [0,1,0], k = 1 Output: 2 Explanation: Flip nums[0], then flip nums[2]. Example 2: Input: nums = [1,1,0], k = 2 Output: -1 Explanation: No matter how we flip subarrays of size 2, we cannot make the array become [1,1,1]. Example 3: Input: nums = [0,0,0,1,0,1,1,0], k = 3 Output: 3 Explanation: Flip nums[0],nums[1],nums[2]: nums becomes [1,1,1,1,0,1,1,0] Flip nums[4],nums[5],nums[6]: nums becomes [1,1,1,1,1,0,0,0] Flip nums[5],nums[6],nums[7]: nums becomes [1,1,1,1,1,1,1,1] Constraints: 1 <= nums.length <= 105 1 <= k <= nums.length

Explanation

Here's the solution to the k-bit flips problem:

• High-Level Approach: The core idea is to iterate through the array and, for each element, check if it's a 0. If it is, we need to flip a subarray of length k starting at that index. To efficiently track flips without repeatedly modifying the array, we use a difference array (or, more precisely, track active flips). We maintain a counter of active flips and update it as we move through the array.

• Complexity: The time complexity is O(n), and the space complexity is O(1).

Code

    def minKBitFlips(nums, k):
    n = len(nums)
    flips = 0
    active_flips = 0
    flipped = [0] * n  # Tracks if a flip starts at this index

    for i in range(n):
        if i >= k and flipped[i - k] == 1:
            active_flips -= 1

        if (nums[i] + active_flips) % 2 == 0:  # Need to flip
            if i + k > n:
                return -1
            flips += 1
            active_flips += 1
            flipped[i] = 1

    return flips