Leftmost Column with at Least a One

Leftmost Column with at Least a One - Problem

You are given a row-sorted binary matrix where all elements are either 0 or 1, and each row is sorted in non-decreasing order.

Your task is to find the leftmost column that contains at least one 1. Return the 0-indexed column number, or -1 if no such column exists.

Important constraints:

You cannot access the matrix directly
You can only use the BinaryMatrix interface with methods:
- get(row, col) - returns the element at position (row, col)
- dimensions() - returns [rows, cols] as a list
Your solution must make at most 1000 calls to BinaryMatrix.get()

Input & Output

Example 1 — Basic Case

$ Input: binaryMatrix = [[0,0],[1,1]]

› Output: 0

💡 Note: The leftmost column with a 1 is column 0. Row 1 has a 1 at position (1,0).

Example 2 — No Ones

$ Input: binaryMatrix = [[0,0],[0,0]]

› Output: -1

💡 Note: There are no 1s in the matrix, so return -1.

Example 3 — Multiple Rows

$ Input: binaryMatrix = [[0,0,1],[0,1,1],[0,0,1]]

› Output: 1

💡 Note: The leftmost column with a 1 is column 1. Row 1 has the first 1 at position (1,1).

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

Visualization

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Asked in

f Facebook 8 G Google 5 a Amazon 3

The key insight is to leverage the sorted property of rows to avoid checking every cell. The optimal approach starts from the top-right corner and moves strategically: left when finding a 1 (to find earlier columns), down when finding a 0 (since no 1s exist to the left in that row). Time: O(m + n), Space: O(1).

Common Approaches

✓ Two Pointers

⏱️ Time: N/A Space: N/A

Binary Search on Each Row

⏱️ Time: O(m × log n) Space: O(1)

Since each row is sorted, we can use binary search to find the leftmost 1 in each row. Track the minimum column index found across all rows.

Brute Force - Column by Column

⏱️ Time: O(m × n) Space: O(1)

Iterate through each column from left to right, and for each column, check all rows to see if any contains a 1. Return the first column that has at least one 1.

Start from Top-Right Corner

⏱️ Time: O(m + n) Space: O(1)

Begin from the top-right corner of the matrix. When we see a 1, move left to find a potentially earlier column. When we see a 0, move down since this row cannot have 1s to the left. This eliminates rows/columns efficiently.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Ln 1, Col 1

Smart Actions

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