minimum-number-of-k-consecutive-bit-flips

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Algorithms

You are given a binary array <code>nums</code> and an integer <code>k</code>.

A k-bit flip is choosing a subarray of length <code>k</code> from <code>nums</code> and simultaneously changing every <code>0</code> in the subarray to <code>1</code>, and every <code>1</code> in the subarray to <code>0</code>.

Return the minimum number of k-bit flips required so that there is no <code>0</code> in the array. If it is not possible, return <code>-1</code>.

A subarray is a contiguous part of an array.

&nbsp;
Example 1:

<pre>
Input: nums = [0,1,0], k = 1
Output: 2
Explanation: Flip nums[0], then flip nums[2].
</pre>

Example 2:

<pre>
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].
</pre>

Example 3:

<pre>
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]
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 105</code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>

minKBitFlips

Minimum Number of K Consecutive Bit Flips

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灯泡开关

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Minimum Operations to Make Binary Array Elements Equal to One I

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Prefix Sum

Sliding Window

给定一个二进制数组 <code>nums</code> 和一个整数 <code>k</code> 。

k位翻转 就是从 <code>nums</code> 中选择一个长度为 <code>k</code> 的 子数组 ，同时把子数组中的每一个 <code>0</code> 都改成 <code>1</code> ，把子数组中的每一个 <code>1</code> 都改成 <code>0</code> 。

返回数组中不存在 <code>0</code> 所需的最小 k位翻转 次数。如果不可能，则返回&nbsp;<code>-1</code>&nbsp;。

子数组 是数组的 连续 部分。

&nbsp;

示例 1：

<pre>
输入：nums = [0,1,0], K = 1
输出：2
解释：先翻转 A[0]，然后翻转 A[2]。
</pre>

示例 2：

<pre>
输入：nums = [1,1,0], K = 2
输出：-1
解释：无论我们怎样翻转大小为 2 的子数组，我们都不能使数组变为 [1,1,1]。
</pre>

示例 3：

<pre>
输入：nums = [0,0,0,1,0,1,1,0], K = 3
输出：3
解释：
翻转 A[0],A[1],A[2]:&nbsp;A变成 [1,1,1,1,0,1,1,0]
翻转 A[4],A[5],A[6]:&nbsp;A变成 [1,1,1,1,1,0,0,0]
翻转 A[5],A[6],A[7]:&nbsp;A变成 [1,1,1,1,1,1,1,1]
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums.length &lt;= 105</code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>