995. Minimum Number of K Consecutive Bit Flips

Hard

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

Solution:

class Solution {

    /**
     * @param Integer[] $nums
     * @param Integer $k
     * @return Integer
     */
    function minKBitFlips($nums, $k) {
        $flipped = array_fill(0, count($nums), false);
        $validFlipsFromPastWindow = 0;
        $flipCount = 0;

        for ($i = 0; $i < count($nums); $i++) {
            if ($i >= $k) {
                if ($flipped[$i - $k]) {
                    $validFlipsFromPastWindow--;
                }
            }
            if ($validFlipsFromPastWindow % 2 == $nums[$i]) {
                if ($i + $k > count($nums)) {
                    return -1;
                }
                $validFlipsFromPastWindow++;
                $flipped[$i] = true;
                $flipCount++;
            }
        }

        return $flipCount;
    }
}

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