leetcode - 1428. Leftmost Column with at Least a One

Description

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:

Input: mat = [[0,0],[1,1]]
Output: 0
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Example 2:

Input: mat = [[0,0],[0,1]]
Output: 1
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Example 3:

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

Binary Search

For each row, do a binary search, and update the leftmost column index.

Time complexity: $o(\log n\ast m)$
Space complexity: $o(1)$

Linear

Solved after hints.

Start from top right, if the current number is 0, go down. Otherwise go left.

Time complexity: $o(m+n)$
Space complexity: $o(1)$

Code

Binary Search

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

class Solution:
    def leftMostColumnWithOne(self, binaryMatrix: 'BinaryMatrix') -> int:
        rows, cols = binaryMatrix.dimensions()
        def find_index(row: int) -> int:
            left, right = 0, cols - 1
            while left < right:
                mid = (left + right) >> 1
                if binaryMatrix.get(row, mid) == 0:
                    left = mid + 1
                else:
                    right = mid
            return (left + right) >> 1
        res = -1
        for i in range(rows):
            col_index = find_index(i)
            if binaryMatrix.get(i, col_index) == 1:
                res = min(res, col_index) if res != -1 else col_index
        return res
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Linear

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

class Solution:
    def leftMostColumnWithOne(self, binaryMatrix: 'BinaryMatrix') -> int:
        rows, cols = binaryMatrix.dimensions()
        cur_row, cur_col = 0, cols - 1
        res = cols
        while cur_row < rows and cur_col >= 0:
            if binaryMatrix.get(cur_row, cur_col) == 1:
                res = min(res, cur_col)
                cur_col -= 1
            else:
                cur_row += 1
        return res if res != cols else -1
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