<p>Given a binary array <code>nums</code>, return <em>the maximum number of consecutive </em><code>1</code><em>&#39;s in the array</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,0,1,1,1]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The first two digits or the last three digits are consecutive 1s. The maximum number of consecutive 1s is 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,0,1,1,0,1]
<strong>Output:</strong> 2
</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 findMaxConsecutiveOnes(vector<int>& nums) {
        stack<int>st;
        int count =0;
        int n = nums.size();
        int max_count =0;
        if(nums[0]== 1)
        {
            count =1;
        }
        if(nums[0]== 1 && n==1)
        {
            return 1;
        }
        st.push(nums[0]);
        for(int i=1; i<n; i++)
        {
            if(nums[i]==1 && st.top()==1 || nums[i]== 1 && st.top()==0)
            {
                count++;
                max_count = max(max_count, count);
            }
            else if(nums[i] == 0)
            {
                max_count = max(max_count, count);
                count =0;
            }
            st.push(nums[i]);
        }

        return max_count;
    }
};
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