LC: 1428. Leftmost Column with at Least a One
https://leetcode.com/problems/leftmost-column-with-at-least-a-one/
1428. Leftmost Column with at Least a One
(This problem is an interactive problem.)
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 isrows 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:

Example 2:

Example 3:

Example 4:

Constraints:
rows == mat.lengthcols == mat[i].length1 <= rows, cols <= 100mat[i][j]is either0or1.mat[i]is sorted in non-decreasing order.
Leftmost Column with at Least a One
The Essence:
Dass die Matrix “row-sorted” ist, weist darauf hin, dass sich die linkste Eins immer rechts von einem Null befindet.
Details:
Die Essenz impliziert 2 Lösungswege:
Der Problemlöser kann die binäre Suche in jeder Zeile verwenden, um das erste Vorkommen von 1 zu finden und die Ergebnisse zu vergleichen.
Wenn die linkste 1 bei einem Index in einer Zeile gefunden wird, müssen nur Indizes unterhalb dieses Indexes in den darunter liegenden Zeilen gesucht werden. Dies kann dafür genutzt werden, um die Suche in den nächsten Zeilen effizienter durchzuführen.
Solution(s) and Further Explanation:
Default Code:
Last updated