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1428. Leftmost Column with at Least a One

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(x, y) returns the element of the matrix at index (x, y) (0-indexed). BinaryMatrix.dimensions() returns a list of 2 elements [m, n], which means the matrix is m * n. 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.

def leftMostColumnWithOne(self, binaryMatrix: 'BinaryMatrix') -> int: 
        
    rows, cols = binaryMatrix.dimensions() 
        
    # Set pointers to the top-right corner. 
    current_row = 0 
    current_col = cols - 1 
        
    # Repeat the search until it goes off the grid. 
    while current_row < rows and current_col >= 0: 
        if binaryMatrix.get(current_row, current_col) == 0: 
            current_row += 1 
        else: 
            current_col -= 1 
        
    # If we never left the last column, it must have been all 0's. 
    return current_col + 1 if current_col != cols - 1 else -1