74. Search a 2D Matrix

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 the last integer of the previous row.

Example 1:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true

Example 2:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false

bool searchMatrix(vector<vector<int>>& matrix, int target) {
    // treat the matrix as an array, just taking care of indices
    // [0..n*m]
    // (row, col) -> row*n + col
    // i -> [i/n][i%n]
    if(matrix.empty() || matrix[0].empty())
    {
        return false;
    }
    int m = matrix.size(), n = matrix[0].size();
    int start = 0, end = m*n - 1;
    while(start <= end)
    {
        int mid = start + (end - start)/2;
        int e = matrix[mid/n][mid%n];
        if(target < e)
        {
            end = mid - 1;
        }
        else if(target > e)
        {
            start = mid + 1;
        }
        else
        {
            return true;
        }
    }
    return false;
}