<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) {
        int noOfZeroes=0;
        int noOfOnes=0;
        for(int i=0;i<nums.size();i++){
            if(nums[i]==0){
                nums[i]=-1;
                noOfZeroes++;
            }
            else{
                noOfOnes++;
            }
        }
        if(noOfZeroes==noOfOnes){
            return nums.size();
        }
        //now we have the updated nums array
        //we need to the longest subarray with sum equal to 0
        unordered_map<int,int>prefixSum;
        prefixSum[0]=-1;
        int sum=0;
        int maxSize=0;
        for(int i=0;i<nums.size();i++){
            sum+=nums[i];
            if(prefixSum.find(sum)!=prefixSum.end()){
                // auto p=prefixSum[sum-0];
                maxSize=max(maxSize,i-prefixSum[sum]);

            }
            if(prefixSum.find(sum)==prefixSum.end()){
            prefixSum[sum]=i;
            }

        }
        return maxSize;
    }
};
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