<p>You are given an integer <strong>mountain</strong> array <code>arr</code> of length <code>n</code> where the values increase to a <strong>peak element</strong> and then decrease.</p>

<p>Return the index of the peak element.</p>

<p>Your task is to solve it in <code>O(log(n))</code> time complexity.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">arr = [0,1,0]</span></p>

<p><strong>Output:</strong> <span class="example-io">1</span></p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">arr = [0,2,1,0]</span></p>

<p><strong>Output:</strong> <span class="example-io">1</span></p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">arr = [0,10,5,2]</span></p>

<p><strong>Output:</strong> <span class="example-io">1</span></p>
</div>

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

<ul>
	<li><code>3 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= arr[i] &lt;= 10<sup>6</sup></code></li>
	<li><code>arr</code> is <strong>guaranteed</strong> to be a mountain array.</li>
</ul>

class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        l, r = 0, len(arr) - 1

        while l <= r:
            mid = (l + r) // 2 

            leftGreater = (mid == 0 or arr[mid] > arr[mid - 1])
            rightGreater = (mid == len(arr) - 1 or arr[mid] > arr[mid + 1])

            if leftGreater and rightGreater:
                return mid 

            if not rightGreater:
                l = mid
            else:
                r = mid

        return -1
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