304 North Cardinal St.
Dorchester Center, MA 02124

# Search a 2D Matrix II LeetCode Solution

## Problem – 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`

### Search a 2D Matrix II LeetCode Solution in Java

``````public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix == null || matrix.length < 1 || matrix.length <1) {
return false;
}
int col = matrix.length-1;
int row = 0;
while(col >= 0 && row <= matrix.length-1) {
if(target == matrix[row][col]) {
return true;
} else if(target < matrix[row][col]) {
col--;
} else if(target > matrix[row][col]) {
row++;
}
}
return false;
}
}
``````

### Search a 2D Matrix II LeetCode Solution in C++

``````class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int m = matrix.size(), n = m ? matrix.size() : 0, r = 0, c = n - 1;
while (r < m && c >= 0) {
if (matrix[r][c] == target) {
return true;
}
matrix[r][c] > target ? c-- : r++;
}
return false;
}
};
``````

### Search a 2D Matrix II LeetCode Solution in Python

``````class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int)-> bool:

if not len(matrix) or not len(matrix):
# Quick response for empty matrix
return False

h, w = len(matrix), len(matrix)

for row in matrix:

# range check
if row <= target <= row[-1]:

# launch binary search on current possible row

left, right = 0, w-1

while left <= right:

mid = left + (right - left) // 2

mid_value = row[mid]

if target > mid_value:
left = mid+1
elif target < mid_value:
right = mid-1
else:
return True

return False
``````
##### Search a 2D Matrix II LeetCode Solution Review:

In our experience, we suggest you solve this Search a 2D Matrix II LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Search a 2D Matrix II LeetCode Solution

Find on LeetCode

##### Conclusion:

I hope this Search a 2D Matrix II LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

More Coding Solutions >>

LeetCode Solutions

Hacker Rank Solutions

CodeChef Solutions