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

Integers in each row are sorted from left to right.
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.
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
boolean found=false;
if(matrix==null){
return found;
}
int m=matrix.length;
int n=matrix[0].length; int upRow=0;
int downRow=m-1; while(upRow<=downRow){
if(target<=matrix[upRow][n-1]){
found=binarySearch(matrix, upRow, target);
break;
}
else if(target>=matrix[downRow][0]){
found=binarySearch(matrix, downRow, target);
break;
}
else{
upRow++;
downRow--;
}
}
return found;
} public boolean binarySearch(int[][]matrix, int row, int target){
int m=matrix.length;
int n=matrix[0].length; int left=0;
int right=n-1; boolean found=false;
while(left<=right){
int mid=left+(right-left)/2;
if(matrix[row][mid]==target){
found=true;
break;
}
else if(matrix[row][mid]<target){
left=mid+1;
}
else{
right=mid-1;
}
}
return found;
}
}

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