Search a Row-Major Sorted Matrix

Given an integer matrix matrix and an integer target, return true when target appears in the matrix and false otherwise.
Each row is sorted in strictly increasing order. The first value of every row after the first is greater than the final value of the previous row.
Use a binary-search solution with logarithmic running time in the total number of cells.

Function
searchMatrix(matrix: int[][], target: int) → boolean

Examples
Example 1
matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]]
target = 3
return = true
The value 3 is the second cell of the first row.
Example 2
matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]]
target = 13
return = false
Binary search ends between 11 and 16, so 13 is absent.
Example 3
matrix = []
target = 1
return = false
An empty matrix contains no target.

Constraints
0 <= matrix.length <= 100.
When matrix is non-empty, 1 <= matrix[i].length <= 100 and every row has the same length.
-10^9 <= matrix[i][j], target <= 10^9.
Rows satisfy the stated global ordering.

Treat the matrix as a flattened sorted array of length rows times cols. Binary search lo = 0, hi = rows*cols - 1. For each mid, row = mid / cols, col = mid % cols, then compare matrix[row][col] to target and move the bounds. That's O(log(m*n)) time and O(1) space. The pitfalls are small but they bite. An empty matrix like Example 3 has no columns, so check matrix.length == 0 before reading matrix[0].length. Mixing up the divisor (use cols, not rows) gives wrong cells on non-square grids. Use lo + (hi - lo) / 2 if your language overflows, though these bounds are small. Don't scan every row, and don't run BFS. If the index conversion slips away under pressure, StealthCoder can surface the clean solution in the live OA so you can verify your own version against it.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.