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
class Solution {
public:
    int findMaxLength(vector<int>& nums) {
        unordered_map<int, int> d{{0, -1}};
        int ans = 0, s = 0;
        for (int i = 0; i < nums.size(); ++i) {
            s += nums[i] ? 1 : -1;
            if (d.contains(s)) {
                ans = max(ans, i - d[s]);
            } else {
                d[s] = i;
            }
        }
        return ans;
    }
};