Leftmost Column with at Least a One

Problem Statement

In computational problems involving matrices, a row-sorted binary matrix is a common type, where each row consists only of 0s and 1s, sorted in non-decreasing order (i.e., all 0s come before any 1s in a row). Given such a binary matrix, which you cannot access directly except through a provided interface BinaryMatrix, the challenge is to find the index of the leftmost column that contains at least one 1.

The BinaryMatrix interface offers:

• BinaryMatrix.get(row, col): Fetches the element at the specified row and column.
• BinaryMatrix.dimensions(): Returns the dimensions of the matrix as a two-element list [rows, cols].

Constraints include a limit of 1000 calls to BinaryMatrix.get to avoid brute force approaches. If no column with a 1 is found, the function should return -1.

Examples

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

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.

Approach and Intuition

Given the unique constraints and format of the problem, an efficient approach is crucial. Here's a step-by-step breakdown:

1. Get Dimensions: Use BinaryMatrix.dimensions() to retrieve the number of rows and columns.

2. Start at Top-Right Corner:

   • The intuition for starting at the top-right is it allows you to eliminate either a row or a column in each step, thus optimizing the search.
   • If you find a 1, you can move left (to reduce the column index) because there can't be a 1 in any column to the right of the current one within the same row.
   • If you find a 0, you should move down to the next row since there can't be a 1 to the left in the same row, but there might be below in the same column.

3. Iterate until Boundaries are Reached:

   • Continue the process until you either run out of columns or rows. If you exit the loop with a column index in the valid range, that's the leftmost column with a 1. If not, return -1.

This strategy ensures that each step provides maximal information, either confirming the absence of 1's to the left or above, allowing us to make the least number of calls to BinaryMatrix.get while unraveling the matrix effectively.

Examples Explained

For Example 1:

• Starting at the top-right (0, 1) gives 0, so move down.
• At (1, 1), it's 1, indicating the potential leftmost column, so move left.
• At (1, 0), it's 1 again - confirming column 0 is indeed the leftmost with a 1.

For Example 2:

• Start at (0, 1), move down because it's 0.
• At (1, 1), find 1 - confirming column 1 is the leftmost with a 1 as moving left leads to (1, 0), which is 0.

For Example 3:

• Checks confirm only 0s present, returning -1.

This approach leverages the matrix's sorted property and bounded interface calls to efficiently locate the desired column index.

Solutions

class Solution {
    public int firstColumnWithOne(BinaryMatrix matrix) {
        int numRows = matrix.dimensions().get(0);
        int numCols = matrix.dimensions().get(1);

        // Start from the top-right corner of the matrix
        int row = 0;
        int col = numCols - 1;
    
        // Move towards the bottom-left corner
        while (row < numRows && col >= 0) {
            if (matrix.get(row, col) == 0) {
                row++;
            } else {
                col--;
            }
        }
    
        // Check if the column is the last one and full of zeros
        return (col == numCols - 1) ? -1 : col + 1;
    }
}

class Solution:
    def findLeftmostOne(self, grid: 'BinaryMatrix') -> int:
        # Retrieve the dimensions of the grid
        num_rows, num_cols = grid.dimensions()

        # Initialize the position as top-right corner
        row_index = 0
        col_index = num_cols - 1

        # Traverse the grid until bounds are exceeded
        while row_index < num_rows and col_index >= 0:
            if grid.get(row_index, col_index) == 0:
                row_index += 1
            else:
                col_index -= 1

        # Check the position where we stopped to determine if a 1 was found
        return col_index + 1 if col_index != num_cols - 1 else -1