<p>Given an integer array <code>nums</code> that <strong>does not contain</strong> any zeros, find <strong>the largest positive</strong> integer <code>k</code> such that <code>-k</code> also exists in the array.</p>

<p>Return <em>the positive integer </em><code>k</code>. If there is no such integer, return <code>-1</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [-1,2,-3,3]
<strong>Output:</strong> 3
<strong>Explanation:</strong> 3 is the only valid k we can find in the array.
</pre>

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

<pre>
<strong>Input:</strong> nums = [-1,10,6,7,-7,1]
<strong>Output:</strong> 7
<strong>Explanation:</strong> Both 1 and 7 have their corresponding negative values in the array. 7 has a larger value.
</pre>

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

<pre>
<strong>Input:</strong> nums = [-10,8,6,7,-2,-3]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There is no a single valid k, we return -1.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>-1000 &lt;= nums[i] &lt;= 1000</code></li>
	<li><code>nums[i] != 0</code></li>
</ul>

def bs(a, l, r, x):
    res = -1
    while l <= r:
        m = (l + r) // 2
        if a[m] == x:
            return abs(a[m])
        elif a[m] > x:
            r = m - 1
        else:
            l = m + 1
    return res

class Solution:
    def findMaxK(self, nums: list[int]) -> int:
        res = -1
        nums.sort()  # Sort the array first
        for i in range(len(nums)):
            if nums[i] > 0:  # Only search for positive numbers
                res = max(res, bs(nums, 0, i-1, -nums[i]))  # Search for -nums[i] in the range before nums[i]
        return res

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