<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>

class Solution
{
public:
  int minKBitFlips(vector<int>& nums, int k)
  {
    int ans{ 0 };
    const auto iEnd{ nums.end() - k + 1 };
    queue<vector<int>::iterator> fep; // Flip End Position

    for(auto i{ nums.begin() }; i != iEnd; i++) {
      if(fep.front() == i) {
        fep.pop();
      }
      if(*i == fep.size() % 2) {
        ans++;
        fep.push(i + k);
      }
      // Set current value of i to it's value after flips
      if(fep.size() % 2 == 1) {
        *i = !*i;
      }
#if MY_DEBUG
      if(nums.size() < 20)
        cout << nums << endl;
#endif
    }

    for(auto i{ iEnd }; i != nums.end(); i++) {
      if(fep.front() == i) {
        fep.pop();
      }
      if(*i == fep.size() % 2) // i==0 and no flip or i==1 and 1 flip
        return -1;
    }

    return ans;
  }
};

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