<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) {
        #define MAX(a, b) ((a) > (b) ? (a) : (b))
        int n(nums.size());
        int sum2idx[200001] = {0};
        fill(sum2idx, sum2idx+200000, -1);
        partial_sum(nums.cbegin(), nums.cend(), nums.begin());
        int maxLen(0);
        for (int i = 0; i < n; i ++) {
            nums[i] = 2 * nums[i] - i - 1;
            int x = nums[i];
            if (x == 0) {
                maxLen = MAX(maxLen, i+1);
            } else {
                if (sum2idx[x+100000] != -1) {
                    maxLen = MAX(maxLen, i - sum2idx[x+100000]);
                } else {
                    sum2idx[x+100000] = i;
                }
            }
        }
        return maxLen;
    }
};
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