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# Search a 2D Matrix II

Updated: Mar 25, 2021

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The 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 <= matix[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

Solution:

Iterative:

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

Recursive:

```class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
return helper(matrix,target,matrix.length-1,0);
}

public boolean helper(int[][] matrix,int target,int i,int j)
{
if(i<0 || j<0 || i>matrix.length-1 || j>matrix[0].length-1) return false;
if(matrix[i][j]==target) return true;
return (matrix[i][j]>target)? helper(matrix,target,i-1,j):helper(matrix,target,i,j+1);
}
}

```

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