<p>A <strong>peak</strong> element in a 2D grid is an element that is <strong>strictly greater</strong> than all of its <strong>adjacent </strong>neighbors to the left, right, top, and bottom.</p>

<p>Given a <strong>0-indexed</strong> <code>m x n</code> matrix <code>mat</code> where <strong>no two adjacent cells are equal</strong>, find <strong>any</strong> peak element <code>mat[i][j]</code> and return <em>the length 2 array </em><code>[i,j]</code>.</p>

<p>You may assume that the entire matrix is surrounded by an <strong>outer perimeter</strong> with the value <code>-1</code> in each cell.</p>

<p>You must write an algorithm that runs in <code>O(m log(n))</code> or <code>O(n log(m))</code> time.</p>

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

<p><img alt="" src="https://assets.leetcode.com/uploads/2021/06/08/1.png" style="width: 206px; height: 209px;" /></p>

<pre>
<strong>Input:</strong> mat = [[1,4],[3,2]]
<strong>Output:</strong> [0,1]
<strong>Explanation:</strong>&nbsp;Both 3 and 4 are peak elements so [1,0] and [0,1] are both acceptable answers.
</pre>

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

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2021/06/07/3.png" style="width: 254px; height: 257px;" /></strong></p>

<pre>
<strong>Input:</strong> mat = [[10,20,15],[21,30,14],[7,16,32]]
<strong>Output:</strong> [1,1]
<strong>Explanation:</strong>&nbsp;Both 30 and 32 are peak elements so [1,1] and [2,2] are both acceptable answers.
</pre>

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

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 500</code></li>
	<li><code>1 &lt;= mat[i][j] &lt;= 10<sup>5</sup></code></li>
	<li>No two adjacent cells are equal.</li>
</ul>

class Solution:
    def max_index(self,mat,mid):
        ind=-1
        value=-1
        for i in range(len(mat)):
            if mat[i][mid]>value:
                value=mat[i][mid]
                ind=i
        return ind
    def findPeakGrid(self, mat: List[List[int]]) -> List[int]:
        low=0
        high=len(mat[0])

        while(low<=high):
            mid=(low+high)//2
            g=self.max_index(mat,mid)
            if mid-1>=0:
                left=mat[g][mid-1]
            else:
                left=-1
            if mid+1<len(mat[0]):
                right=mat[g][mid+1]
            else:
                right=-1
            if (mat[g][mid]>left and mat[g][mid]>right):
                return [g,mid]
            elif mat[g][mid]<left:
                high=mid-1
            else:
                low=mid+1

        
883