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Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has properties:

1) Integers in each row are sorted from left to right. 2) The first integer of each row is greater than the last integer of the previous row.

For example, consider the following matrix:

```
[
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
```

Given target = 3, return true.

**Java Solution**

This is a typical problem of binary search.

You may try to solve this problem by finding the row first and then the column. There is no need to do that. Because of the matrix’s special features, the matrix can be considered as a sorted array. The goal is to find the element in this sorted array by using binary search.

```
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix==null || matrix.length==0 || matrix[0].length==0)
return false;
int m = matrix.length;
int n = matrix[0].length;
int start = 0;
int end = m*n-1;
while(start<=end){
int mid=(start+end)/2;
int midX=mid/n;
int midY=mid%n;
if(matrix[midX][midY]==target)
return true;
if(matrix[midX][midY]<target){
start=mid+1;
}else{
end=mid-1;
}
}
return false;
}
}
```