Contiguous Array
<p>Given a binary array <code>nums</code>, return <em>the maximum length of a contiguous subarray with an equal number of </em><code>0</code><em> and </em><code>1</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [0,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> [0, 1] is the longest contiguous subarray with an equal number of 0 and 1.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,1,0]
<strong>Output:</strong> 2
<strong>Explanation:</strong> [0, 1] (or [1, 0]) is a longest contiguous subarray with equal number of 0 and 1.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,1,1,1,1,1,0,0,0]
<strong>Output:</strong> 6
<strong>Explanation:</strong> [1,1,1,0,0,0] is the longest contiguous subarray with equal number of 0 and 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>nums[i]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>
Array; Hash Table; Prefix Sum
func findMaxLength(nums []int) int {
	d := map[int]int{0: -1}
	ans, s := 0, 0
	for i, x := range nums {
		if x == 0 {
			x = -1
		}
		s += x
		if j, ok := d[s]; ok {
			ans = max(ans, i-j)
		} else {
			d[s] = i
		}
	}
	return ans
}