Minimum Swaps to Group All 1's Together
<p>Given a&nbsp;binary array <code>data</code>, return&nbsp;the minimum number of swaps required to group all <code>1</code>&rsquo;s present in the array together in <strong>any place</strong> in the array.</p>

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

<pre>
<strong>Input:</strong> data = [1,0,1,0,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong> There are 3 ways to group all 1&#39;s together:
[1,1,1,0,0] using 1 swap.
[0,1,1,1,0] using 2 swaps.
[0,0,1,1,1] using 1 swap.
The minimum is 1.
</pre>

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

<pre>
<strong>Input:</strong> data = [0,0,0,1,0]
<strong>Output:</strong> 0
<strong>Explanation:</strong> Since there is only one 1 in the array, no swaps are needed.
</pre>

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

<pre>
<strong>Input:</strong> data = [1,0,1,0,1,0,0,1,1,0,1]
<strong>Output:</strong> 3
<strong>Explanation:</strong> One possible solution that uses 3 swaps is [0,0,0,0,0,1,1,1,1,1,1].
</pre>

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

<ul>
	<li><code>1 &lt;= data.length &lt;= 10<sup>5</sup></code></li>
	<li><code>data[i]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>
Array; Sliding Window
class Solution {
    public int minSwaps(int[] data) {
        int k = 0;
        for (int v : data) {
            k += v;
        }
        int t = 0;
        for (int i = 0; i < k; ++i) {
            t += data[i];
        }
        int mx = t;
        for (int i = k; i < data.length; ++i) {
            t += data[i];
            t -= data[i - k];
            mx = Math.max(mx, t);
        }
        return k - mx;
    }
}