<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>&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>

class Solution {
public:
    int findMaxLength(vector<int>& nums) {
        // Edge case
        if (nums.size() <= 1) return 0;

        // Lenghth of subarray
        int length = 0;

        // Tracking vector
        unordered_map<int, pair<int, int>> tracker;
        tracker[0] = {0, 0};

        int num = 0;
        
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] == 0) num++;
            else num--;
            if (tracker.find(num) == tracker.end()) {
                tracker[num] = {i+1, 0};
            }
            else {
                tracker[num].second = i+1;
                length = max(length, tracker[num].second - tracker[num].first);
            }
        }

        return length;
    }
};
89300