## 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.

For 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.

## Analysis:

Since the matrix is sorted by column and row, we can easliy know that:
• matrix[ i ] [ j ]  <= matrix [ i-1 ] [ j ]
• matrix[ i ] [ j ]  >= matrix [ i ] [ j-1 ]
For the element in the top right (or bottom left), we could find out:
• if target < top right, then target cannot be in last column.
• if target > top right, then target cannot be in the first row.
Therefore, search from the top right (or bottom left) with two pointers (for row and column), will eliminate either a row or a column in each iteration, the overall time complexity is O(m + n).

## Code (C++):

class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int nRow = matrix.size();
if (nRow == 0) {return false;}
int nCol = matrix[0].size();
if (nCol == 0) {return false;}
int i = 0;
int j = nCol - 1;
while (i < nRow && j >=0){
if (matrix[i][j] < target){
i++;
}else if (matrix[i][j] > target){
j--;
}else{
return true;
}
}
return false;
}
};

## Code (Python):

class Solution(object):
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
i = 0
j = len(matrix[0]) -1
while i < len(matrix) and j >= 0:
if matrix[i][j] < target:
i += 1
elif matrix[i][j] > target:
j -= 1
else:
return True
return False