[LC] 240. Search a 2D Matrix II

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 in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

Example:

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

Solution 1: 

Time: O(M * N)

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return false;
        }
        for (int[] arr: matrix) {
            if (binarySearch(arr, target)) {
                return true;
            }
        }
        return false;
    }
    
    private boolean binarySearch(int[] array, int target) {
        int start = 0, end = array.length - 1;
        while (start <= end) {
            int mid = start + (end - start) / 2;
            if (array[mid] == target) {
                return true;
            } else if (array[mid] < target) {
                start = mid + 1;
            } else {
                end = mid - 1;
            }
        }
        return false;
    }
}

Solution 2:

Time: O(M + N)

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return false;
        }
        // check from upper right corner
        int row = 0, col = matrix[0].length - 1;
        while (col >= 0 && row <= matrix.length - 1) {
            if (matrix[row][col] == target) {
                return true;
            } else if (matrix[row][col] < target) {
                row += 1;
            } else {
                col -= 1;
            }
        }
        return false;
    }
}