# Search a 2D Matrix II LeetCode Solution | Easy Approach Share:

Search a 2D Matrix II Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `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 1:

```Input: 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]], target = 5
Output: true
```

Example 2:

```Input: 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]], target = 20
Output: false
```

Constraints:

• `m == matrix.length`
• `n == matrix[i].length`
• `1 <= n, m <= 300`
• `-109 <= matrix[i][j] <= 109`
• All the integers in each row are sorted in ascending order.
• All the integers in each column are sorted in ascending order.
• `-109 <= target <= 109`

Time: O(m+n)
Space: O(1)

### C++

`````` class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int r = 0;
int c = matrix.size() - 1;

while (r < matrix.size() && c >= 0) {
if (matrix[r][c] == target)
return true;
if (matrix[r][c] > target)
--c;
else
++r;
}

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

### Java

`````` class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
int r = 0;
int c = matrix.length - 1;

while (r <= matrix.length && c >= 0) {
if (matrix[r][c] == target)
return true;
if (matrix[r][c] > target)
--c;
else
++r;
}

return false;
}
}

``````

### Python

`````` class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
r = 0
c = len(matrix) - 1

while r < len(matrix) and c >= 0:
if matrix[r][c] == target:
return True
if target < matrix[r][c]:
c -= 1
else:
r += 1

return False

``````

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