Leftmost Column with at Least a One

https://leetcode.com/explore/featured/card/30-day-leetcoding-challenge/530/week-3/3306/

A binary matrix means that all elements are 0 or 1. For each individual row of the matrix, this row is sorted in non-decreasing order.

Given a row-sorted binary matrix binaryMatrix, return leftmost column index(0-indexed) with at least a 1 in it. If such index doesn't 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 a list of 2 elements [rows, cols], which means the matrix is rows * 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 you're given the binary matrix mat as input in the following four examples. You will not have access the binary matrix directly.

Example 1:

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

Example 2:

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

Example 3:

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

Example 4:

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 a non-decreasing way.

01矩阵每行已经排好序，问从左边数第一个出现1的列是多少。让指针指向第一行最后一个元素，如果是1往前走，否则往下走，依此类推。因为已经排好序，当遇到0意味着这一行的1肯定不可能出现在上一行1前面的位置，此行可以排除。

# """
# This is BinaryMatrix's API interface.
# You should not implement it, or speculate about its implementation
# """
#class BinaryMatrix(object):
#    def get(self, x: int, y: int) -> int:
#    def dimensions(self) -> list[]:

class Solution:
    def leftMostColumnWithOne(self, binaryMatrix: 'BinaryMatrix') -> int:
        M, N = binaryMatrix.dimensions()
        r, c = 0, N - 1
        while r < M and c >= 0:
            if binaryMatrix.get(r, c) == 0:
                r += 1
            else:
                c -= 1
        return c + 1 if c != N - 1 else -1