# Binary Matrix Leftmost One

You are given a two-dimensional list of integers `matrix`

which contains `1`

s and `0`

s. Given that each row is sorted in ascending order with `0`

s coming before `1`

s, return the leftmost column index with the value of `1`

. If there’s no row with a `1`

, return `-1`

.

Can you solve it faster than \(\mathcal{O}(nm)\).

**Constraints**

`n, m ≤ 250`

where`n`

and`m`

are the number of rows and columns in`matrix`

https://binarysearch.com/problems/Binary-Matrix-Leftmost-One

## Examples

### Example 1

**Input**

- matrix =

```
[[0,0,0,0],
[0,0,1,1],
[0,0,0,1],
[0,1,1,1]]
```

**Output**

- answer =
`1`

**Explanation**

The last row contains the leftmost column with a one at index `1`

.

## Leave a comment