<p>You are given an <code>m x n</code> integer matrix <code>matrix</code> with the following two properties:</p>

<ul>
	<li>Each row is sorted in non-decreasing order.</li>
	<li>The first integer of each row is greater than the last integer of the previous row.</li>
</ul>

<p>Given an integer <code>target</code>, return <code>true</code> <em>if</em> <code>target</code> <em>is in</em> <code>matrix</code> <em>or</em> <code>false</code> <em>otherwise</em>.</p>

<p>You must write a solution in <code>O(log(m * n))</code> time complexity.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/05/mat.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/05/mat2.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
<strong>Output:</strong> false
</pre>

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

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
	<li><code>-10<sup>4</sup> &lt;= matrix[i][j], target &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        m = len(matrix)
        if m == 0:
            return False
        n = len(matrix[0])

        l , r = 0 , m*n - 1 
        while l <= r:
            m = (l + r )//2 
            value = matrix[m // n][ m % n]
            if target == value:
                return True
            else:
                if target < value:
                    r = m - 1 
                else:
                    l = m + 1 
        return False

        
        
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