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

<p>A <strong>k-bit flip</strong> is choosing a <strong>subarray</strong> 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>.</p>

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

<p>A <strong>subarray</strong> is a <strong>contiguous</strong> part of an array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,0], k = 2
<strong>Output:</strong> -1
<strong>Explanation:</strong> No matter how we flip subarrays of size 2, we cannot make the array become [1,1,1].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,0,0,1,0,1,1,0], k = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> 
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>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>

// Greedy
/*
There is only one way to filp A[0],
and A[0] will tell us if we need to filp the range A[0] ~ A[K -1].
So we start from the leftmost one by one using a greedy idea to solve this problem.

Create a new array isFlipped[n].
isFlipped[i] = 1 iff we flip K consecutive bits starting at A[i].

We maintain a variable flipped and flipped = 1 iff the current bit is flipped.

If flipped = 0 and A[i] = 0, we need to flip at A[i].
If flipped = 1 and A[i] = 1, we need to flip at A[i].

Complexity
O(N) time for one pass
O(N) extra space for isFlipped[n].

if we use mod, only store the sliding windows size, then we can reduce space to O(K).

There is a constant space solution, too tricky, don't even look at it.
*/
class Solution {
public:
    int minKBitFlips(vector<int>& nums, int k) {
        int n = nums.size();
        int flipped = 0;
        vector<int> isFlipped(n, 0);

        int res = 0;
        for (int i = 0; i < n; ++i)
        {
            if (i >= k)
            {
                flipped ^= isFlipped[i-k];  //抵消上一个k sliding window的flip影响，因为影响不到当前元素了
            }

            if (flipped == nums[i]) // flipped == 0 && nums[i] == 0 或者 二者都是1. Needs flip
            {
                if (i + k > n) return -1;  // no more window to flip
                isFlipped[i] = 1;
                flipped ^= 1;

                res++;  // needs another flip
            }
        }

        return res;
    }
};
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