<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) {
        unordered_map<int, int> ct;
        int ans = 0;
        int csum = 0;

        for(int i=0; i<nums.size(); i++){
            int val = nums[i] == 0 ? -1 : 1;
            csum += val;

            if(csum == 0){
                ans = max(ans, i+1);
            }
            else if(ct.find(csum) != ct.end()){
                //pehle se hi present hai
                ans = max(ans, i - ct[csum]);
            }
            else{
                //absent..so create new key with value same as index of that element
                ct[csum] = i;
            }
        }
        return ans;
    }
};
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