Minimum Number of K Consecutive Bit Flips

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 <= 10^5
• 1 <= k <= nums.length

Solutions

Approach

To solve this problem, we can use a greedy approach with a sliding window to keep track of the flips. The main idea is to flip the bits only when necessary and use an auxiliary array to record the effect of flips at each position.

1. Initialize Variables:

   • flipped to track the cumulative effect of flips.
   • res to count the number of flips.
   • isFlipped array to record the flips at each index.

2. Iterate through the Array:

   • For each element, check if the current position is affected by a previous flip by using the isFlipped array.
   • If the current bit needs to be flipped (flipped == nums[i]), check if flipping is possible.
   • Record the flip and update the counters.

3. Edge Cases:

   • If it's not possible to flip because the subarray exceeds the array bounds, return -1.

Java

class Solution {
    public int minKBitFlips(int[] nums, int k) {
        int n = nums.length, flipped = 0, res = 0;
        int[] isFlipped = new int[n];
        
        for (int i = 0; i < nums.length; ++i) {
            if (i >= k)
                flipped ^= isFlipped[i - k];
            if (flipped == nums[i]) {
                if (i + k > nums.length)
                    return -1;
                isFlipped[i] = 1;
                flipped ^= 1;
                res++;
            }
        }
        return res;
    }
}

Python

class Solution:
    def minKBitFlips(self, nums: List[int], k: int) -> int:
        n = len(nums)
        flipped = 0
        res = 0
        is_flipped = [0] * n
        
        for i in range(n):
            if i >= k:
                flipped ^= is_flipped[i - k]
            if flipped == nums[i]:
                if i + k > n:
                    return -1
                is_flipped[i] = 1
                flipped ^= 1
                res += 1
        
        return res