Search a 2D Matrix
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0074.Search%20a%202D%20Matrix/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0074.Search%20a%202D%20Matrix/images/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>
Array; Binary Search; Matrix
/**
 * @param {number[][]} matrix
 * @param {number} target
 * @return {boolean}
 */
var searchMatrix = function (matrix, target) {
    const m = matrix.length,
        n = matrix[0].length;
    let left = 0,
        right = m * n - 1;
    while (left < right) {
        const mid = (left + right + 1) >> 1;
        const x = Math.floor(mid / n);
        const y = mid % n;
        if (matrix[x][y] <= target) {
            left = mid;
        } else {
            right = mid - 1;
        }
    }
    return matrix[Math.floor(left / n)][left % n] == target;
};
