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
func minSwaps(data []int) int {
	k := 0
	for _, v := range data {
		k += v
	}
	t := 0
	for _, v := range data[:k] {
		t += v
	}
	mx := t
	for i := k; i < len(data); i++ {
		t += data[i]
		t -= data[i-k]
		mx = max(mx, t)
	}
	return k - mx
}