Binary search of a Matrix

Question

Write an efficient algorithm that searches for a value in an m x n matrix.

This matrix has the following properties:

-Integers in each row are sorted from left to right.
-The first integer of each row is greater than or equal to the last integer of the previous row.

Example:

Consider the following matrix:

[
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
Given target = 3, return 1 ( 1 corresponds to true )

Return 0 / 1 ( 0 if the element is not present, 1 if the element is present ) for this problem

My solution works on NetBeans but fails on the website. Any help will be appreciated. https://www.interviewbit.com/problems/matrix-search/

public class Solution {

    public int searchMatrix(ArrayList<ArrayList<Integer>> a, int b) {
        int r = a.size();
        int c = a.get(0).size();

        int start = 0;
        int end = r - 1;
        // default value is last row for edge case
        int biRow = r -1; // row to search column

        //binary search 1st value of rows
        while (start <= end) {
            int mid = (start + end) / 2;
            if (b == a.get(mid).get(0)) {
                return 1;
            }

            if (a.get(mid).get(0) < b && b < a.get(end).get(0)) {
                if (mid + 1 >= end) {
                    biRow = mid;
                    break;
                }
            } if (b < a.get(mid).get(0)) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        }
        //binary search column of biRow
        start = 0;
        end = c-1;
        while (start <= end) {
            int mid = (start + end) / 2;
            if (b == a.get(biRow).get(mid)) {
                return 1;
            }
            if (b < a.get(biRow).get(mid)) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        }
        return 0;
    }
}

Answer 1

I think the shared program seems to have a logical error.

When updating the end value in the first while loop, if the end value is equal to start, biRow can not be updated.

It worked well when I updated the code like below.

public class Solution {
    public int searchMatrix(ArrayList<ArrayList<Integer>> a, int b) {
        int r = a.size();
        int c = a.get(0).size();
        int start = 0;
        int end = r - 1;
        // default value is last row for edge case
        int biRow = r -1; // row to search column

        //binary search 1st value of rows
        int mid = 0;
        while (start <= end) {
            mid = (start + end) / 2;
            if ( b >= a.get(mid).get(0) && b <= a.get(mid).get(c-1)) {
                break;
            }
            if (b < a.get(mid).get(0)) {
                end = mid-1;
            } else {
                start = mid+1;
            }
       }

        biRow = mid;

        //binary search column of biRow
        start = 0;
        end = c-1;
        while (start <= end) {
            mid = (start + end) / 2;
            if (b == a.get(biRow).get(mid)) {
                return 1;
            }
            if (b < a.get(biRow).get(mid)) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        }
        return 0;
    }
}