<p>Given an array <code>nums</code> sorted in <strong>non-decreasing</strong> order, return <em>the maximum between the number of positive integers and the number of negative integers.</em></p>

<ul>
	<li>In other words, if the number of positive integers in <code>nums</code> is <code>pos</code> and the number of negative integers is <code>neg</code>, then return the maximum of <code>pos</code> and <code>neg</code>.</li>
</ul>

<p><strong>Note</strong> that <code>0</code> is neither positive nor negative.</p>

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

<pre>
<strong>Input:</strong> nums = [-2,-1,-1,1,2,3]
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are 3 positive integers and 3 negative integers. The maximum count among them is 3.
</pre>

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

<pre>
<strong>Input:</strong> nums = [-3,-2,-1,0,0,1,2]
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are 2 positive integers and 3 negative integers. The maximum count among them is 3.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,20,66,1314]
<strong>Output:</strong> 4
<strong>Explanation:</strong> There are 4 positive integers and 0 negative integers. The maximum count among them is 4.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2000</code></li>
	<li><code>-2000 &lt;= nums[i] &lt;= 2000</code></li>
	<li><code>nums</code> is sorted in a <strong>non-decreasing order</strong>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Can you solve the problem in <code>O(log(n))</code> time complexity?</p>

class Solution:
    def maximumCount(self, nums: List[int]) -> int:
        dict = Counter(nums)
        l =0
        r = len(nums) -1 
        ans = 0
        res = 0
        while l <= r:
            mid = (l+r) // 2
            if nums[mid] < 0:
                l = mid +1 
                ans = mid +1
            else:
                r = mid -1
        
        res = max(ans,(len(nums) - (dict[0] + ans)))
        return res 
         
            




        
        
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