# LeetCode 1428. Leftmost Column with at Least a One

*(This problem is an **interactive problem**.)*

A **row-sorted binary matrix** means that all elements are `0` or `1` and each row of the matrix is sorted in non-decreasing order.

Given a **row-sorted binary matrix** `binaryMatrix`, return *the index (0-indexed) of the **leftmost column** with a 1 in it*. If such an index does not exist, return `-1`.

**You can't access the Binary Matrix directly.** You may only access the matrix using a `BinaryMatrix`interface:

* `BinaryMatrix.get(row, col)` returns the element of the matrix at index `(row, col)` (0-indexed).
* `BinaryMatrix.dimensions()` returns the dimensions of the matrix as a list of 2 elements `[rows, cols]`, which means the matrix is `rows x cols`.

Submissions making more than `1000` calls to `BinaryMatrix.get` will be judged *Wrong Answer*. Also, any solutions that attempt to circumvent the judge will result in disqualification.

For custom testing purposes, the input will be the entire binary matrix `mat`. You will not have access to the binary matrix directly.

**Example 1:**

![](https://assets.leetcode.com/uploads/2019/10/25/untitled-diagram-5.jpg)

```
Input: mat = [[0,0],[1,1]]
Output: 0
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2019/10/25/untitled-diagram-4.jpg)

```
Input: mat = [[0,0],[0,1]]
Output: 1
```

**Example 3:**

![](https://assets.leetcode.com/uploads/2019/10/25/untitled-diagram-3.jpg)

```
Input: mat = [[0,0],[0,0]]
Output: -1
```

**Example 4:**

![](https://assets.leetcode.com/uploads/2019/10/25/untitled-diagram-6.jpg)

```
Input: mat = [[0,0,0,1],[0,0,1,1],[0,1,1,1]]
Output: 1
```

**Constraints:**

* `rows == mat.length`
* `cols == mat[i].length`
* `1 <= rows, cols <= 100`
* `mat[i][j]` is either `0` or `1`.
* `mat[i]` is sorted in non-decreasing order.

## Solution:

[English Version in Youtube](https://youtu.be/O3KQIh0J89g)

[中文版解答Youtube Link](https://youtu.be/koAFGCKT-Hw)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1ef4y1478d/)

```
/**
 * // This is the BinaryMatrix's API interface.
 * // You should not implement it, or speculate about its implementation
 * class BinaryMatrix {
 *   public:
 *     int get(int row, int col);
 *     vector<int> dimensions();
 * };
 */

class Solution {
public:
    int leftMostColumnWithOne(BinaryMatrix &binaryMatrix) {
        vector<int> dimension = binaryMatrix.dimensions();
        int row = dimension[0];
        int col = dimension[1];
        int res = -1;
        for (int i = 0; i < row; i++) {
            int left = 0;
            int right = (res == -1 ? col - 1 : res - 1);
            while (left < right) {
                int mid = left + (right - left) / 2;
                if (binaryMatrix.get(i, mid) == 0) {
                    left = mid + 1;
                } else {
                    right = mid;
                }
            }

            if (binaryMatrix.get(i, left) == 1) {
                res = left;
            }
        }
        return res;
    }
};
```


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