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 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

Example 2:

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

Example 3:

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

Example 4:

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

分析：
因为我们只知道每个row是sorted，但是colo并没有。最直接的办法是对于每个row, 返回是1的最小column值， 然后每个row比较一下就知道了。

class Solution {
    public int leftMostColumnWithOne(BinaryMatrix binaryMatrix) {
        List<Integer> dim = binaryMatrix.dimensions();
        int r = dim.get(0), c = dim.get(1);
        int leftMost = c;
        
        for (int i = 0; i < r; i++) {
            leftMost = Math.min(leftMost, leftMostBinarySearch(binaryMatrix, i, c, 1));
        }
        return leftMost == c ? -1 : leftMost;
    }
    
    private int leftMostBinarySearch(BinaryMatrix binaryMatrix, int row, int colos, int target) {
        int left = 0;
        int right = colos - 1; 
        while (left <= right) {
            int mid = left + (right - left)/2;
            if (binaryMatrix.get(row, mid) < 1) {
                left = mid+1;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }

还有一种更好的方法。我们从第一行最后一个column开始，如果值是1，我们就往左走，这样能够知道第一行第一个1的index,然后我们继续往下，check有没有比当前1更靠左的index。

    public int leftMostColumnWithOne(BinaryMatrix binaryMatrix) {
        List<Integer> dimensions = binaryMatrix.dimensions();
        int rows = dimensions.get(0), cols = dimensions.get(1);
        int ans = -1, row = 0, col = cols - 1;
        while (row < rows && col >= 0) {
            if (binaryMatrix.get(row, col) == 1) {
                ans = col; 
                col--;
            } else {
                row++;
            }
        }
        return ans;
    }

Reference: https://leetcode.com/problems/leftmost-column-with-at-least-a-one/discuss/590828/Java-Binary-Search-and-Linear-Solutions-with-Picture-Explain-Clean-Code