Input: mat = [[0,0,0,1],[0,0,1,1],[0,1,1,1]] Output: 1
A binary matrix means that all elements are
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
You can't access the Binary Matrix directly. You may only access the matrix using a
BinaryMatrix.get(x, y)returns the element of the matrix at index
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.
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.
- (Binary Search) For each row do a binary search to find the leftmost one on that row and update the answer.
- (Optimal Approach) Imagine there is a pointer p(x, y) starting from top right corner. p can only move left or down. If the value at p is 0, move down. If the value at p is 1, move left. Try to figure out the correctness and time complexity of this algorithm.
# https://leetcode.com/explore/challenge/card/30-day-leetcoding-challenge/530/week-3/3306/ # """ # 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() column = -1 # Initiate initial row to show last seen row row = 0 # start and end to show starting and ending # column values that we want to search start = 0 end = n # Break condition for binary search while(start != end): middle = start + (end - start) // 2 middle_el = binaryMatrix.get(row, middle) # if middle element is 1 then move left in the row # and set end point to middle of the row if(middle_el == 1): end = middle column = middle # if middle element is 0 then check middle elements # of the remaining rows else: found1 = False for i in range(row+1, m): el = binaryMatrix.get(i, middle) # if 1 found in any row set that row as starting row if(el == 1): row = i end = middle column = middle found1 = True break if (not found1): start = middle+1 return column
Problem descripion and image source: leetcode