Searching a 2D Sorted Matrix II | LeetCode

Searching a 2D Sorted Matrix Part II | LeetCode
Write an efficient algorithm that searches for a value in an n x m table (two-dimensional array). This table is sorted along the rows and columns — that is,
Table[i][j] ≤ Table[i][j + 1],
Table[i][j] ≤ Table[i + 1][j]

Step-wise Linear Search - O(N)We call this the Step-wise Linear Search method. Similar to Diagonal Binary Search, we begin with the upper right corner (or the bottom left corner). Instead of traversing diagonally each step, we traverse one step to the left or bottom.

Essentially, each step we are able to eliminate either a row or a column. The worst case scenario is where it ended up in the opposite corner of the matrix, which takes at most 2n steps. Therefore, this algorithm runs in O(n) time

Java code from EPI
  public static boolean matrixSearch(List<List<Integer>> A, int x) {
    int r = 0, c = A.get(0).size() - 1;
    while (r < A.size() && c >= 0) {
      if (A.get(r).get(c).equals(x)) {
        return true;
      } else if (A.get(r).get(c) < x) {
        ++r;
      } else { // A.get(r).get(c) > x.
        --c;
      }
    }
    return false;
  }
Quad Partition:

T(n)  = 3lg n ( T(1) + c ) - c/2 
       = O(3lg n) 
       = O(nlg 3)           <== 3lg n = nlg 3 
       = O(n1.58)

bool quadPart(int mat[][N_MAX], int M, int N, int target, int l, int u, int r, int d, int &targetRow, int &targetCol) {

  if (l > r || u > d) return false;

  if (target < mat[u][l] || target > mat[d][r]) return false;

  int col = l + (r-l)/2;

  int row = u + (d-u)/2;

  if (mat[row][col] == target) {

    targetRow = row;

    targetCol = col;

    return true;

  } else if (l == r && u == d) {

    return false;

  }

  if (mat[row][col] > target) {

    return quadPart(mat, M, N, target, col+1, u, r, row, targetRow, targetCol) ||

           quadPart(mat, M, N, target, l, row+1, col, d, targetRow, targetCol) ||

           quadPart(mat, M, N, target, l, u, col, row, targetRow, targetCol);

  } else {

    return quadPart(mat, M, N, target, col+1, u, r, row, targetRow, targetCol) ||

           quadPart(mat, M, N, target, l, row+1, col, d, targetRow, targetCol) ||

           quadPart(mat, M, N, target, col+1, row+1, r, d, targetRow, targetCol);

  }

}

bool quadPart(int mat[][N_MAX], int N, int target, int &row, int &col) {

  return quadPart(mat, N, N, target, 0, 0, N-1, N-1, row, col);

}

Improved Binary Partition: 

 T(n) = 2T(n/2) + c lg n
                       lg n
      = 2 lg n T(1) + c ∑ ( 2 lg (n/2k) ) 
                       k=1
      = O(n) + c (2n - lg n - 2)
      = O(n)

bool binPart(int mat[][N_MAX], int M, int N, int target, int l, int u, int r, int d, int &targetRow, int &targetCol) {

  if (l > r || u > d) return false;

  if (target < mat[u][l] || target > mat[d][r]) return false;

  int mid = l + (r-l)/2;

  int row = u;

  while (row <= d && mat[row][mid] <= target) {

    if (mat[row][mid] == target) {

      targetRow = row;

      targetCol = mid;

      return true;

    }

    row++;

  }

  return binPart(mat, M, N, target, mid+1, u, r, row-1, targetRow, targetCol) ||

         binPart(mat, M, N, target, l, row, mid-1, d, targetRow, targetCol);

}

bool binPart(int mat[][N_MAX], int N, int target, int &row, int &col) {

  return binPart(mat, N, N, target, 0, 0, N-1, N-1, row, col);

}

Question:
2D Matrix(n * n) of positive and negative numbers is given. Matrix is sorted rowwise and columnwise. You have to return the count of -ve numbers in most optimal way.
Also check http://leetcode.com/2010/10/searching-2d-sorted-matrix.html
http://leetcode.com/2010/10/searching-2d-sorted-matrix-part-iii.html
https://www.youtube.com/watch?t=348&v=0-rX-Wocuew

http://www.zrzahid.com/search-in-a-2d-array-or-matrix/

Version 1
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.

1. If the mid is greater than the key than we know that this row and all next rows do not contain the mid. So, search upward. 

2. On the other hand if mid is less than the key than either this row may contain the key or all subsequent rows. 

Looks suspicious:

public boolean searchMatrix1(int[][] matrix, int target) {
    if(matrix == null || matrix.length == 0){
        return false;
    }
    
    //find the row
    int n = matrix.length;
    int m = matrix[0].length;
    
    int row = -1;
    int rl = 0;
    int rh = n-1;
    while(rl <= rh){
        int mid = (rl+rh)/2;
        if(matrix[mid][0] == target || matrix[mid][m-1] == target){
            return true;
        }
        else if(matrix[mid][0] > target){
            rh = mid-1;
        }
        else{
            if(matrix[mid][m-1] > target){
                row = mid;
                break;
            }
            else{
                rl = mid+1;
            }
        }
    }
    
    if(row == -1){
        return false;
    }
    
    //search in the row
    int col = -1;
    int cl = 0;
    int ch = m-1;
    while(cl <= ch){
        int mid = (cl+ch)/2;
        if(matrix[row][mid] == target){
            return true;
        }
        else if(matrix[row][mid] > target){
            ch = mid-1;
        }
        else{
            cl = mid+1;
        }
    }
    
    return false;
}

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