## Search a 2D Matrix

• Time:O(mn\log mn)
• Space:O(1)

## C++

``````class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if (matrix.empty())
return false;

const int m = matrix.size();
const int n = matrix[0].size();
int l = 0;
int r = m * n;

while (l < r) {
const int mid = (l + r) / 2;
const int i = mid / n;
const int j = mid % n;
if (matrix[i][j] == target)
return true;
if (matrix[i][j] < target)
l = mid + 1;
else
r = mid;
}

return false;
}
};
``````

## JAVA

``````class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix.length == 0)
return false;

final int m = matrix.length;
final int n = matrix[0].length;
int l = 0;
int r = m * n;

while (l < r) {
final int mid = (l + r) / 2;
final int i = mid / n;
final int j = mid % n;
if (matrix[i][j] == target)
return true;
if (matrix[i][j] < target)
l = mid + 1;
else
r = mid;
}

return false;
}
}
``````

## Python

``````class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
if not matrix:
return False

m = len(matrix)
n = len(matrix[0])
l = 0
r = m * n

while l < r:
mid = (l + r) // 2
i = mid // n
j = mid % n
if matrix[i][j] == target:
return True
if matrix[i][j] < target:
l = mid + 1
else:
r = mid

return False
``````